Math Analysis Homework - Week 2

Posted on 2025-09-26 Edited on 2025-09-29

Math Analysis Homework - Week 2

Class 4

T1

设 $x_{1} > 1, x_{n + 1} = \frac{3 x_{n} + 1}{x_{n} + 3}, n = 1, 2, \dots$

\begin{matrix} x_{n} > 1 \Rightarrow 1 < x_{n + 1} = \frac{3 x_{n} + 1}{x_{n} + 3} < 3 \\ Inductively, x_{n} \in (1, 3) x_{n + 1} - x_{n} \\ = \frac{3 x_{n} + 1 - x_{n}^{2} - 3 x_{n}}{x_{n} + 3} \\ = \frac{1 - x_{n}^{2}}{x_{n} + 3} < 0 \\ ∴ {x_{n}} is bounded and decreasing \\ lim_{n \to \infty} x_{n} exists . \\ x_{n + 1} = \frac{3 x_{n} + 1}{x_{n} + 3} \\ \Rightarrow lim_{n \to \infty} x_{n} = \frac{3 (lim_{n \to \infty} x_{n}) + 1}{(lim_{n \to \infty} x_{n}) + 3} \\ \Rightarrow lim_{n \to \infty} x_{n} = 1 \end{matrix}

T2

设 $0 < a_{1} < b_{1}$ , 令 $a_{n + 1} = \sqrt{a_{n} b_{n}}, b_{n + 1} = \frac{a_{n} + b_{n}}{2}, n \in N_{+}$ . 证明: ${a_{n}}, {b_{n}}$ 收敛于同一极限.

\begin{matrix} \forall n, a_{n} < b_{n} \\ \Rightarrow {\begin{cases} b_{n} > a_{n + 1} = \sqrt{a_{n} b_{n}} > a_{n} \\ a_{n} < b_{n + 1} = \frac{a_{n} + b_{n}}{2} < b_{n} \end{cases} \\ \Rightarrow {\begin{cases} a_{n} < b_{n} < b_{n - 1} < \dots < b_{1}, \\ b_{n} > a_{n} > a_{n - 1} > \dots > a_{1} \end{cases} \\ \Rightarrow {a_{n}}, {b_{n}} is bounded and monotonic \\ \Rightarrow A = lim_{n \to \infty} a_{n}, B = lim_{n \to \infty} b_{n} exists \\ \Rightarrow {\begin{cases} lim_{n \to \infty} b_{n + 1} = lim_{n \to \infty} \frac{a_{n} + b_{n}}{2} \\ lim_{n \to \infty} a_{n + 1} = \sqrt{a_{n} b_{n}} \end{cases} \\ \Rightarrow {\begin{cases} B = \frac{A + B}{2} \\ A = \sqrt{A B} \end{cases} \Rightarrow A = B \end{matrix}

T3

设数列 ${x_{n}}$ 满足 $0 < x_{n} < 1$ 与 $(1 - x_{n}) x_{n + 1} > \frac{1}{4}, n = 1, 2, 3, \dots$ , 求证: $lim_{n \to \infty} x_{n} = \frac{1}{2}$ .

\begin{matrix} \frac{1}{4 (1 - x_{n})} < x_{n + 1} < 1 \\ \Rightarrow x_{n} < \frac{3}{4} \\ \frac{1}{4 (1 - x_{n - 1})} < x_{n} < \frac{3}{4} \\ \Rightarrow x_{n - 1} < \frac{2}{3} \\ 同理 \Rightarrow x_{n - 2} < \frac{1}{2} \\ ∴ \forall n, x_{n} < \frac{1}{2} \\ 又 x_{n + 1} = \frac{1}{4 (1 - x_{n})} > x_{n} \\ \Leftrightarrow (2 x_{n} - 1)^{2} > 0 \Rightarrow True \\ X = lim_{n \to \infty} x_{n} exists \\ x_{n + 1} > \frac{1}{4 (1 - x_{n})} \\ \Rightarrow X \geq \frac{1}{4 (1 - X)} \\ \Rightarrow X = \frac{1}{2} \end{matrix}

T4

设 $x_{1} \in (0, 1), x_{n + 1} = x_{n} (1 - x_{n}), n \in N_{+}$ , 证明: 数列 ${n x_{n}}$ 收敛, 并求其极限.

\begin{matrix} lim_{n \to \infty} \frac{\frac{1}{x_{n}}}{n} \\ \overset{Stolz Theorem}{⟸} lim_{n \to \infty} \frac{1}{x_{n}} - \frac{1}{x_{n - 1}} \\ = lim_{n \to \infty} \frac{1}{x_{n - 1} (1 - x_{n - 1})} - \frac{1}{x_{n - 1}} \\ = lim_{n \to \infty} \frac{1}{1 - x_{n - 1}} \\ = \frac{1}{1 - lim_{n \to \infty} x_{n - 1}} \end{matrix}

对于 $x$ ,

\begin{matrix} \begin{array}{ll} x_{1} \leq \frac{1}{1}, x_{2} < min (x_{1}, 1 - x_{1}) \leq \frac{1}{2} \\ x_{n} \leq \frac{1}{n}, n \geq 2 \Rightarrow x_{n + 1} < \frac{1}{n} (1 - \frac{1}{n}) = \frac{n - 1}{n^{2}} < \frac{1}{n + 1} \end{array}} \\ \Rightarrow x_{n} \leq \frac{1}{n} \\ 0 < x_{n} \leq \frac{1}{n} \\ \overset{Squeeze Theorem}{⟹} \\ lim_{n \to \infty} x_{n} = 0 \\ \Rightarrow lim_{n \to \infty} \frac{\frac{1}{x_{n}}}{n} = \frac{1}{1 - lim_{n \to \infty} x_{n}} = 1 \\ \Rightarrow lim_{n \to \infty} n x_{n} = 1 \end{matrix}

T5

求极限 $lim_{n \to \infty} (n! e - [n! e])$ .

\begin{matrix} a_{n} = n! e - [n! e] \\ = lim_{n \to \infty} {n! e} \\ = lim_{n \to \infty} {\sum_{i = 0}^{\infty} \frac{n!}{i!}} \\ = lim_{n \to \infty} {\sum_{i = n + 1}^{\infty} \frac{n!}{i!}} \\ = lim_{n \to \infty} {\frac{1}{n + 1} + \frac{1}{(n + 1) (n + 2)} + \frac{1}{(n + 1) (n + 2) (n + 3)} + \dots} \\ < lim_{n \to \infty} {\sum_{i = 1}^{\infty} (n + 1)^{- i}} \\ = lim_{n \to \infty} {\frac{1}{n}} \\ = 0 \end{matrix}

T6

求极限 $lim_{n \to \infty} a_{n} (\frac{1}{n + 1} + \frac{1}{n + 2} + \dots + \frac{1}{n + n})$ .

\begin{matrix} x > \ln (x) + 1 \\ \Rightarrow \frac{1}{x} \in (\ln (\frac{x}{x - 1}), \ln (\frac{x + 1}{x})) \\ \Rightarrow a_{n} \in (\sum_{i = n + 1}^{2 n} \ln (\frac{i}{i - 1}), \sum_{i = n + 1}^{2 n} \ln (\frac{i + 1}{i})) \\ = (\ln (\frac{2 n}{n}), \ln (\frac{2 n + 1}{n + 1})) \\ \overset{Squeeze Theorem}{⟹} lim_{n \to \infty} a_{n} = \ln (2) \end{matrix}

T7

设数列 $x_{n} = (1 + \frac{1}{2}) (1 + \frac{1}{2^{2}}) \dots (1 + \frac{1}{2^{n}})$ , 证明 $lim_{n \to \infty} x_{n}$ 存在.

\begin{matrix} x_{n} is obviously incresing \\ \ln (x_{n}) = \sum_{i = 1}^{n} \ln (1 + \frac{1}{2^{i}}) \\ < \sum_{i = 1}^{n} \frac{1}{2^{i}} \\ < 1 \\ \Rightarrow x_{n} < e \Rightarrow lim_{n \to \infty} x_{n} exists \end{matrix}

Class 5

T1

用柯西收敛准则证明数列收敛.

\begin{matrix} a_{n} = \sum_{i = 2}^{n} \frac{\sin (i x)}{i (i + \sin (i x))}, x \in R \end{matrix}

\begin{matrix} | a_{n + m} - a_{n} | \\ = | \sum_{i = n + 1}^{n + m} \frac{\sin (i x)}{i (i + \sin (i x))} | \\ < \sum_{i = n + 1}^{n + m} | \frac{\sin (i x)}{i (i + \sin (i x))} | \\ < \sum_{i = n + 1}^{n + m} \frac{1}{i (i - 1)} \\ = \sum_{i = n + 1}^{n + m} \frac{1}{i - 1} - \frac{1}{i} \\ = \frac{1}{n} - \frac{1}{n + m} \\ < \frac{1}{n} \\ \Rightarrow \forall ϵ, N := \frac{1}{ϵ} + 100 \\ \Rightarrow \forall i, j > N, | a_{i} - a_{j} | < ϵ \\ \overset{Cauchy Convergence Theorem}{⟹} \\ Q.E.D \end{matrix}

T2

\begin{matrix} b_{n} = \sum_{i = 1}^{n - 1} | a_{i + 1} - a_{i} | is bounded \Rightarrow a_{n} is convergent \end{matrix}

\begin{matrix} \begin{array}{ll} b_{n} is increasing \\ b_{n} is bounded \end{array}} \Rightarrow lim_{n \to \infty} b_{n} = B \\ a_{n + m} - a_{n} = \sum_{i = n + 1}^{n + m} a_{i} - a_{i - 1} \\ \leq \sum_{i = n + 1}^{n + m} | a_{i} - a_{i - 1} | \\ = b_{n + m} - b_{n} \\ \Rightarrow \forall ϵ_{1}, \exists N \\ s . t . \\ n > N \Rightarrow | b_{n} - B | < ϵ_{1} \\ \Rightarrow b_{n + m} - b_{n} < | b_{n} - B | + | B - b_{n + m} | = 2 ϵ_{1} \\ ϵ_{1} := \frac{ϵ}{2} \Rightarrow \forall x, y > N, | a_{x} - a_{y} | < ϵ \\ \overset{Cauchy Convergence Theorem}{⟹} \\ Q.E.D \end{matrix}

T3

\begin{matrix} \forall ϵ, \exists N_{1} = N (ϵ) \\ s . t . \\ i, j > N_{1} \Rightarrow | x_{i} - x_{j} | < ϵ \\ \Leftrightarrow \\ x_{n} is convergent \end{matrix}

逆向三角不等式显然.考虑正向

\begin{matrix} ϵ_{1} := 1 ∴ i > N \Rightarrow x_{i} \in [x_{N} - ϵ_{1}, x_{N} + ϵ_{1}] \\ a_{1} := x_{N} - ϵ_{1}, b_{1} := x_{N} + ϵ_{1} \end{matrix}

对 $[a_{i}, b_{i}]$ ,考虑取 $ϵ_{i} = \frac{ϵ_{i - 1}}{2}$ ,有

N_{i} = max (N_{i - 1}, N (ϵ_{i})) s . t . \forall j > N_{i}, x_{j} \in [x_{N_{i}} - ϵ_{i}, x_{N_{i}} + ϵ_{i}]

于是令 $a_{i} = max (a_{i - 1}, x_{N_{i}} - ϵ), b_{i} = min (b_{i - 1}, x_{N_{i}} + ϵ_{i})$ . 显然 $| b_{i} - a_{i} | < \frac{1}{2^{i - 1}}$ ,于是

\begin{matrix} \begin{array}{ll} a_{i} \leq a_{i + 1} \\ b_{i} \geq b_{i + 1} \\ lim_{n \to \infty} b_{n} - a_{n} = 0 \end{array}} \\ \Rightarrow \exists! ξ \in [a_{n}, b_{n}] \\ s . t . \\ \forall ϵ, N = min {i | b_{i} - a_{i} < ϵ} \Rightarrow n > N \Rightarrow | ξ - x_{n} | < ϵ \\ Q.E.D \end{matrix}

T4

\begin{matrix} a_{0} = 3, a_{n} = a_{n - 1}^{2} - 2 \\ \Rightarrow {\begin{cases} lim_{n \to \infty} a_{n} = + \infty \\ A_{n} := \frac{a_{n}}{\prod_{i = 0}^{n - 1} a_{i}} \Rightarrow lim_{n \to \infty} A_{n} = \sqrt{5} \end{cases} \end{matrix}

(1)

\begin{matrix} \begin{array}{ll} a_{0} = 3, a_{1} = 7, a_{2} = 47 \\ n > 2, a_{n - 1} > 2^{n} \Rightarrow a_{n} = a_{n - 1}^{2} - 2 > 2^{2 n} - 2 > 2^{n + 1} \end{array}} \\ \overset{induction}{⟹} a_{n} > 2^{n + 1} \\ ∴ lim_{n \to \infty} a_{n} \geq lim_{n \to \infty} 2^{n + 1} = \infty \end{matrix}

(2)

\begin{matrix} a_{n} = a_{n - 1}^{2} - 2 \\ \Rightarrow (a_{n} - 2) = (a_{n - 1} - 2) (a_{n - 1} + 2) \\ \Rightarrow \prod_{i = 1}^{n} (a_{i} + 2) = \frac{a_{n + 1} - 2}{a_{1} - 2} = \frac{a_{n + 1} - 2}{5} \\ \Rightarrow \\ A_{n}^{2} = \frac{a_{n}^{2}}{\prod_{i = 0}^{n - 1} a_{i}^{2}} \\ = \frac{a_{n + 1} + 2}{\prod_{i = 1}^{n} (a_{i} + 2)} \\ = 5 \frac{a_{n + 1} + 2}{a_{n + 1} - 2} \\ ∴ lim_{n \to \infty} A_{n} = \sqrt{5 \frac{a_{n + 1} + 2}{a_{n + 1} - 2}} = \sqrt{5} \end{matrix}

Class 6

T1

\begin{matrix} A, B is upper bounder, S \subset {x + y | x \in A, y \in B} \\ \Rightarrow sup S \leq sup A + sup B \end{matrix}

显然 $S$ 有界,则 $sup S$ 存在.

若 $sup S > sup A + sup B$ ,取 $M = sup A + sup B$ ,

则 $\forall x \in A, y \in B, x \leq sup A, y \leq sup B \Rightarrow x + y \leq M$

于是 $M$ 是比 $sup S$ 小的上界,矛盾.

故 $sup S \leq sup A + sup B$

eg. 当 $A, B, S$ 均为有限集恰好为 ${ x+y\vert x\in A,y\in B } $ 时显然严格不等号不成立.

T2

\begin{matrix} \begin{array}{ll} A, B aren’t empty, α \geq 0, \\ C = A + α B = {x | x = a + α b, a \in A, b \in B} \end{array}} \\ \Rightarrow sup (A + α B) = M = sup A + α sup B \end{matrix}

\begin{matrix} \forall c \in C = a + α b, \\ a \leq sup A, b \leq sup B \Rightarrow a + α b \leq sup A + α sup B = M \\ \forall M^{'} < M, ϵ = M - M^{'} \\ let X = sup A - \frac{ϵ}{3}, Y = sup B + \frac{ϵ}{3 α} \\ According to the definition of supremum, \exists a > X \in A, y > Y \in B \\ \Rightarrow \exists c = a + b \in C, c > X + Y > M^{'} \Rightarrow sup C = sup A + α sup B \end{matrix}

T3

\begin{matrix} A \subset B \\ \Rightarrow sup A \leq sup B, inf A \geq inf B \end{matrix}

\begin{matrix} if sup A > sup B \\ let M = sup B \overset{Def of supremum}{⟹} \exists a \in A, a > M = sup B \\ \begin{array}{ll} a \in A \overset{A \subset B}{⟹} a \in B \\ a > sup B \end{array}} \Rightarrow False \\ ∴ sup A \leq sup B \end{matrix}

取 $C = - A, D = - B, C \subset D, sup C \leq sup D \Rightarrow inf A \geq inf B$ ,

T4

\begin{matrix} \forall x \in A, y \in B, x \leq y \\ \Rightarrow sup A \leq inf B \end{matrix}

假设 $sup A > inf B$ , $M \in (sup A, inf B)$ ,由 $sup A$ 定义 $\exists a \in A, a > M$ ,由 $inf B$ 定义 $\exists b \in B, b < M$ ,故 $a > M > b$ ,与 $\forall x \in A, y \in B, x \leq y$ 矛盾.

故 $sup A \leq inf B$

Class 7

T1

\begin{matrix} lim_{x \to x_{0}} \sqrt{x} = {\sqrt{x}}_{0} \end{matrix}

if $x_{0} \neq 0$

\begin{matrix} | \sqrt{x} - \sqrt{x_{0}} | \\ = | \frac{x - x_{0}}{\sqrt{x} + \sqrt{x_{0}}} | \\ \leq \frac{δ}{\sqrt{x_{0}}} \\ ∴ δ := \frac{\sqrt{x_{0}} ϵ}{2} \Rightarrow \\ \forall ϵ, x \in N (x_{0}, δ), | \sqrt{x} - \sqrt{x_{0}} | < ϵ \end{matrix}

if $x_{0} = 0, {\sqrt{x}}_{0} = 0$

\begin{matrix} δ := \frac{ϵ^{2}}{4} \Rightarrow \forall ϵ, x \in N (x_{0}, δ), \\ | \sqrt{x} - \sqrt{x_{0}} | = \frac{ϵ}{4} < ϵ \end{matrix}

\begin{matrix} Q.E.D \end{matrix}

T2

\begin{matrix} lim_{x \to + \infty} (\sqrt{x + 1} - \sqrt{x - 1}) = 0 \end{matrix}

\begin{matrix} \sqrt{x + 1} - \sqrt{x - 1} \\ = \frac{2}{\sqrt{x + 1} + \sqrt{x - 1}} \\ < \frac{1}{\sqrt{x}} \\ ∴ \forall ϵ \in (0, 1), δ := \frac{4}{ϵ^{2}}, \\ x > δ \Rightarrow \sqrt{x + 1} - \sqrt{x - 1} < \frac{1}{\sqrt{x}} = \frac{ϵ}{2} < ϵ \\ Q.E.D \end{matrix}

Linear Algebra Done Right

Posted on 2025-09-25 Edited on 2025-09-29

Linear Algebra Done Right

记录大致讲了什么

向量空间

我们定义向量空间

向量空间

向量空间定义在域 $F$ 上,要求支持:

加法
$F$ 中数的数乘
加法单位元(0元)
加法逆元
加法交换结合
数乘结合
加法对数乘分配
对加法和数乘封闭

加粗部分用于判断子空间,根据 $F$ 区分是否是实向量空间/复向量空间

子空间

就是子集且是向量空间的对吧

判定可以看上面

维数

向量空间还应该有维数.于是定义

线性无关组

Group v_{1} \dots v_{n} \in V is linear independet \Leftrightarrow \forall {c_{n}}, c_{i} \in F, \sum_{i = 1}^{n} c_{i} v_{i} = 0 \Leftrightarrow \forall i, c_{i} = 0

张成,张成组

\begin{matrix} span (v_{1} \dots v_{n}) = {\sum_{i = 1}^{n} c_{i} v_{i}} \end{matrix}

张成 $V$ 的组简称张成组

基

基=张成组+线性无关组

则按照直觉的,维数应该是张成组长度的最小值,线性无关组长度的最大值,基的长度等等,下面讨论的是有限维线性空间.

$\dim V$ 是张成组最小长度
$\dim V$ 是线性无关组组最大长度
$\dim V$ 是任意一组基的长度

首先说明,对一个线性相关的组,我们一定可以去掉一个线性相关项保持张成空间不变(显然).

首先说明

任意一组线性无关组长度小于等于任意一组张成组长度:

考虑一个线性无关组和一个张成组,将一个线性无关组的元素加入张成组,则形成的一定是线性相关组,删去一个张成组中的元素则保持张成,不断重复这个操作,注意因为被加入的线性无关,所以你想相关一定得带张成组中的,于是可以一直操作.

直到线性无关组全部被加入,则因为每次删掉一个张成组元素,你一定有线性无关组长度不大于它.

线性无关组可通过加元素扩展到基,张成组可通过删除元素到基

对线性无关组,每次加入一个不属于张成空间的,由于组的长度大小不断增加,而存在一个长度的基,所以你的过程会停止.

同理对张成组每次删掉一个线性相关项不影响张成空间.

每组基的长度都相当,具有恰当长度的线性无关组/张成组是基.

第一句用小于等于关系,后面两个通过基的长度相等+上一条可以变成基说明.

维数

维数就是这个长度为 $\dim V$

空间的运算

空间的和

\begin{matrix} U + V = {u + v | u \in U, v \in V} \end{matrix}

加法实际上是并(包含子空间所有向量的最小空间).

并定义直和.

空间直和

\begin{matrix} W = U \oplus V = U + V \\ s . t . \\ \forall w \in W, \exist! u \in U, v \in V, u + v = w \end{matrix}

直和相当于不交并,是对空间进行一种分解.

然后直和的判定容易证明只要 $0$ 的表示满足唯一性,并推出当且仅当它们的交只有 $0$

空间的积

就是笛卡尔积.

\begin{matrix} U \times V = {(u, v) | u \in U, v \in V} \end{matrix}

其实也可以先扩展一下直和是等价的.

仿射空间

\begin{matrix} u + V = {u + v | u + V} \\ (u + V) + (w + V) = (u + w) + V \\ λ (u + V) = λ u + V \end{matrix}

注意如果 $u \in V$ 则不变, $u \notin V$ 则是一个没有 $0$ 的空间,在此空间的线性组合变成了 $\sum_{i = 0}^{n} c_{i} v_{i} s . t . \sum_{i = 0}^{n} c_{i} = 1$ .

然后仿射空间关于以上运算是线性空间,其中 $0 = V$

空间的商

\begin{matrix} U / V = {u + V | u \in U} \end{matrix}

可以理解成等价类分类.如果两个向量的差在 $V$ 中则认为他们等价.新的空间中每个元素都是一个等价类.而你仿射空间的变换也可以看成是对代表元做变换.

最后考虑它们的维数,有:

\begin{matrix} \dim U \times V = \dim U \oplus V = \dim U + \dim V \\ \dim U + V = \dim U + \dim V - \dim U \cap V \\ \dim U / V = \dim U - \dim V \end{matrix}

第一行是显然的.

第二行考虑取 $U \cup V$ 的一组基,再其中添加上 $U - V$ 的基和 $V - U$ 的基.

第三行需要线性映射.

线性映射

线性映射基本性质

线性映射

线性映射是映射满足:

齐性: $λ v = λ T v$
加性: $T (u + v) = T u + T v$

线性变换的运算

\begin{matrix} (S + T) u = S u + T u \\ (λ T) u = λ (T u) \\ (S T) u = S (T (u)) \end{matrix}

符合直觉的.于是你可以说明 $L (U, V)$ (由 $T : U \to V$ 组成的集合)是 $\dim U \times \dim V$ 维的线性空间,他的标准基可以是所有把 $U$ 的一个基映到 $V$ 的一个基的映射.

值域,零空间

\begin{matrix} for linear map T : U \to V \\ range T = {v | v = T u, u \in U} \\ null T = {u | T u = 0, u \in U} \end{matrix}

\begin{matrix} \dim U = \dim null T + \dim range T \end{matrix}

考虑 $null T$ 的基 $u_{1} \dots u_{n}$ ,并添加 $v_{1} \dots v_{m}$ 扩充到 $U$ 的基.

考虑 $T v_{1} \dots T v_{m}$ 若线性相关, $\sum_{i = 1}^{m} c_{i} T v_{i} = 0 \Rightarrow T \sum_{i = 1}^{m} c_{i} v_{i} = 0$ ,则 $w = \sum_{i = 1}^{m} c_{i} v_{i} \in null T, w = \sum_{i = 1}^{m} c_{i} v_{i} = \sum_{i = 1}^{n} d_{i} u_{i}$ ,与 $u_{i}, v_{i}$ 构成一组基矛盾.

于是 $T v_{i}$ 线性无关,且容易注意到任意 $w \in U, T w = \sum_{i = 1}^{n} c_{i} T u_{i} + \sum_{i = 1}^{m} d_{i} T v_{i} = \sum_{i = 1}^{m} d_{i} T v_{i} \in span (T v_{1} \dots T V_{m})$ 故得证.

单射,满射,双射,可逆

单射: $T x \neq T y \Rightarrow x \neq y$
满射: $\forall v \in V, \exists u \in U, T u = v$
双射就是同时有两条
对于双射 $T$ 的定义 $T^{- 1}$ 满足 $T T^{- 1} = T^{- 1} T = I$

基本性质

单射等价于只把 $0$ 映射到 $0$ .
存在 $U \to V$ 单射说明 $\dim U \leq \dim V$ .
$U \to V$ 有单射说明 $V \to U$ 有满射.
$U \to V$ 存在双射称为 $U, V$ 同构,可以证明任意向量空间同构于某个 $F^{n}$ .

都挺显然的.

\begin{matrix} \dim U / V = \dim U - \dim V \end{matrix}

定义 $T \in L (U, U / V)$ 为商变换,把 $u$ 映射到 $u + V$ .

则

\begin{matrix} null T = V \\ range T = U / V \end{matrix}

套用上面值域和零空间的维数公式即可.

线性映射的矩阵

矩阵

选取 $U, V$ 分别一组基 $u_{1} \dots u_{m}$ , $v_{1} \dots v_{n}$ ,可以把线性变换 $T$ 写成 $n \times m$ 的矩阵 $M (T, u_{1} \dots u_{m}, v_{1} \dots v_{n}) = A_{n \times m}$ 满足

\begin{matrix} T u_{i} = \sum_{j = 1}^{n} A_{j, i} v_{j} \end{matrix}

即每一列是一个基向量在像空间中的基的表示.然后有的时候也直接简写 $M (T)$ .

矩阵可以这么定义因为线性变换的线性保证了你可以只用基的变换去描述它,同时基的这种变换也能唯一确定线性变换.

于是可以定义矩阵运算:

\begin{matrix} M (S) M (T) = M (S T) \\ M (S) + M (T) = M (S T) \\ λ M (T) = M (λ T) \end{matrix}

其中第一行矩阵乘法用坐标写一下可以推出经典的矩阵乘法方式.

算子,不变子空间,商算子和限制算子

这些概念会在本征值那里用到但从属性上讲和这里线性映射关系更大.

算子

\begin{matrix} T \in L (V, V) \end{matrix}

即映射到自身空间的线性变换.

不变子空间

即对算子 $T$ ,有 $\forall u \in U, T u \in U$ ,则 $U$ 为不变子空间.

商算子,限制算子

\begin{matrix} T_{/ U} \in L (V / U, V / U), (T_{/ U}) (v + U) = (T v + U) \\ T |_{U} \in L (U, U), T |_{U} v = T v \end{matrix}

显然 $T |_{U}$ 要求了 $U$ 是不变子空间.

把算子放到更小的空间去研究的方式.

对偶

线性泛函,对偶空间

线性泛函就是 $f \in L (V, F)$ ,所有这样的 $f$ 组成线线性空间 $V^{'}$ 是 $V$ 的对偶空间.

线性泛函可以看成是向量/点的对偶.线性泛函 ${φ | φ_{i} e_{j} = [i = j]}$ 构成对偶空间的基.

对偶映射

若 $T \in L (U, V)$ ,定义 $T^{'} \in L (V^{'}, U^{'})$ 满足

\begin{matrix} \forall f \in V^{'}, T^{'} f = f T \end{matrix}

$T^{'}$ 是反的可以理解因为 $V^{'}$ 的泛函的输入才是 $T$ 的输出.导致没法根据这几个东西定义一个正的出来.

然后对偶主要解释了: $M (T^{'}) = M (T)^{T}$ (右上角的 $T$ 是转置的意思).

对偶映射的运算

\begin{matrix} (S T)^{'} = T^{'} S^{'} \\ (S + T)^{'} = S^{'} + T^{'} \\ (λ S)^{'} = λ S^{'} \end{matrix}

\begin{matrix} (S T)^{'} f = f S T = T^{'} (f S) = T^{'} (S^{'} f) \\ (S + T)^{'} f = f (S + T) = f S + f T = S^{'} f + T^{'} f \\ (λ S)^{'} f = f λ S = λ f S = S^{'} f \end{matrix}

零化子

对线性空间 $V$ 来说,子空间 $U$ 的零化子 $U^{0} = {f | f \in V^{'}, \forall u \in U, f u = 0}$ .

注意 $U^{0}$ 同时依赖 $U$ 和 $V$ .

\begin{matrix} \dim U + \dim U^{0} = \dim V \end{matrix}

取 $U$ 的基 $u_{1} \dots u_{n}$ 扩充到 $V$ 的基 $u_{1} \dots u_{n}, u_{n + 1} \dots u_{n + m}$ .并取 $V^{'}$ 的标准基 $φ_{i} u_{j} = [i = j]$ ,则显然 $f \in U^{0}$ 要求 $f$ 不能有 $φ_{i}, i < n$ 的分量,而任意 $i > n$ 的分量都可以有.于是得证.

\begin{matrix} null T^{'} = (range T)^{0} \\ range T^{'} = (null T)^{0} \end{matrix}

考虑 $T^{'} f u = f T u = 0$ 关于所有 $u$ 成立,则 $f$ 的范围是什么.看右侧显然是 $(range T)^{0}$ 看左侧则是 $(null T^{'})^{0}$ .于是得证.

对第二行,左边是任意 $T^{'} g = g T$ ,右边说你这个线性泛函把所有 $T u = 0$ 的映到 $0$ ,恰好是左边的 $g T$ 满足条件.于是得证.

然后还有一个问题是我们以为 $T^{″} = T$ ,但实际上你甚至不能保证 $V$ 和 $V^{″}$ 是相同的.然后有个典范同构的概念形容他俩的关系就是存在一种不依赖于基的选取的同构(只要定义 $T (u) f = f u$ ,则 $u$ 到 $T (u)$ 是双射.)

本征值基础

本征值

本征值,本征向量

若对算子 $T$ , $\exists v \neq 0 \in V, λ \in F s . t . T v = λ v$ ,则 $λ, v$ 分别为本征值,本征向量.

就是说算子在这个方向上对变换只有伸缩.

一个本征值可能对应多个线性不相关的本征向量,它们构成本征空间 $E (λ, T)$

$λ$ 是 $T$ 的本征值等价于 $T - λ I$ 不是双射,或不是单射/满射

首先注意到对算子来说单射,满射双射等价

又因为单射等价于 $null T = 0$ 所以 $(T - λ I) v = 0$ 和它不是单射等价.

不同本征值对应对本征向量线性不相关.

反证,你要利用不同本征值这个性质,于是你设 $v_{n} \in span (v_{1} \dots v_{n - 1})$ 且 $n$ 为满足条件对最小的.

\begin{matrix} v_{n} = \sum_{i = 1}^{n - 1} c_{i} v_{i} \\ T v_{n} = \sum_{i = 1}^{n - 1} T c_{i} v_{i} \\ λ_{n} (\sum_{i = 1}^{n - 1} c_{i} v_{i}) = \sum_{i = 1}^{n - 1} λ_{i} c_{i} v_{i} \\ 0 = \sum_{i = 1}^{n - 1} (λ_{i} - λ_{n}) c_{i} v_{i} \end{matrix}

因为 $n$ 是最小的,所以 $v_{1} \dots v_{n - 1}$ 线性无关,然后你就推出矛盾.

有此容易说明本征值个数不大于线性空间维数.

复向量空间中的线性映射一定有本征值

考虑

\begin{matrix} v \in V, \dim V = n \\ v, T v, T^{2} v \dots T^{n} V is dependent \\ \sum_{i = 0}^{n} c_{i} T^{i} v = 0 \\ \overset{代数基本定理}{⟹} (\prod_{i = 1}^{n} (T - λ_{i} I)) v = 0 \Rightarrow \exists i, T - λ_{i} I = 0 \\ \Rightarrow λ_{i} is a eigenvalue of T \end{matrix}

上三角矩阵

按照上面基的理解,有

\begin{matrix} M (T, u_{1} \dots u_{n}) is upper triangular matrix \\ \Leftrightarrow \forall i, T u_{i} \in span (u_{1} \dots u_{i}) \\ \Leftrightarrow \forall i, span (u_{1} \dots u_{i}) is invariant space \end{matrix}

感觉是显然的.

那么考虑什么样的线性映射 $T$ 有一组基 $u_{1} \dots u_{n}$ 有上三角矩阵 $A_{n \times n}$ .

\begin{matrix} \forall T \in L (V, V), V is complex vector space \\ \Rightarrow \exists u_{1} \dots u_{n}, M (T, u_{1} \dots u_{n}) is upper triangular matrix \end{matrix}

Proof 1

归纳,假设对任意维数小于 $\dim V$ 的空间成立,考虑取 $T$ 的任意本征值 $λ$ ,则 $U := range T - λ I$ ,则因为 $T$ 不是单的所以 $\dim U < \dim V$ .且 $\forall u \in U, T u = (T - λ I) u + λ u \in U$ ,所以 $T$ 在 $U$ 不变.

于是可以应用归纳结假设, $T |_{U}$ 在 $U$ 上有一组基 $u_{1} \dots u_{n}$ 使得 $M (T |_{U}, u_{1} \dots u_{n})$ 是上三角矩阵.

将这组基扩展到 $V$ 上成为 $u_{1} \dots u_{n}, v_{1} \dots v_{m}$ ,则对 $\forall i$ , $T v_{i} = (T - λ I) v_{i} + λ v_{i} \in span (u_{1} \dots u_{n}) + span (v_{i}) \subset span (u_{1} \dots u_{n}, v_{1} \dots v_{i})$ ,于是是上三角矩阵.

Proof 2

同样归纳,取任意本征向量 $u, U := span (u)$ ,考虑 $T_{/ U}$ 是维数为 $\dim V - 1$ 的空间 $V / U$ 上算子.则它有上三角矩阵.于是存在 $v_{1} + U \dots v_{n} + U$ ,使得 $\forall v + U \in V / U, T_{/ U} (v + U) \in span (v_{1} + U, \dots, v_{n} + U)$ ,也就有 $T v \in span (v_{1}, \dots, v_{n})$ .

然后现在把 $v_{1} \dots v_{n}, u$ 作为新的基,容易发现 $T u = λ u \in span (v_{1}, \dots, v_{n}, u)$ 满足条件.于是存在上三角矩阵.

都要从维度归纳,第二个自然一点吧:商空间就是抹去若干维度.

[think] 但是第一个从 $T - λ I$ 的值域出发是什么个意思?主要利用两个性质:是不变子空间,以及 $T v = (T - λ I) v + λ v$ .是不是相当于把其他向量也拆的"像"本征向量了.

$T$ 有逆等价于 $T$ 的上三角矩阵对角线全部非 $0$

先假设矩阵有逆,设空间 $V$ 基为 $v_{1} \dots v_{n}$ .

\begin{matrix} T v_{1} = A_{1, 1} v_{1} \Rightarrow A_{1, 1} \neq 0 \\ T v_{k} = u + A_{k, k} v_{k}, u \in span (v_{1} \dots v_{k - 1}) \\ if A_{k, k} = 0 \\ T v_{k} \in span (v_{1} \dots v_{k - 1}) \\ ∵ v_{1} \dots v_{k - 1} is independent \\ ∴ T v_{1} \dots T v_{k - 1} is independent, so it is a base \\ ∴ T v_{k} \in span (T v_{1} \dots T v_{k - 1}) \\ \exists c s . t . \sum_{i = 1}^{k} c_{i} T v_{i} = 0 \\ \overset{T^{- 1}}{⟹} \sum_{i = 1}^{k} c_{i} v_{i} = 0 \\ contradiction! \end{matrix}

再假设 $T$ 关于 $V$ 的基 $v_{1} \dots v_{n}$ 的矩阵为上三角矩阵且对角线元素非 $0$ .

那么我们知道 $T v_{i} = A_{i, i} v_{i} + \sum_{j = 1}^{i - 1} c_{j} T v_{j - 1}$ ,其中后一项属于 $span (v_{1} \dots v_{i - 1})$ ,于是容易发现 $T v_{1} \dots T v_{n}$ 线性独立,是一组基,于是 $T$ 是满的,于是 $T$ 可逆.

$T$ 的某个基下的上三角矩阵对角线元素是 $T$ 的本征值.

考虑 $(T - λ I) v = 0$ ,则 $λ$ 是本征值等价于 $T - λ I$ 不是单的,也就不是可逆的,即用上面条件对角线存在 $0$ ,即 $λ$ 等于对角线上的某个元素.

对角矩阵

本征空间

\begin{matrix} E (λ, T) = null T - λ I \end{matrix}

$T$ 在基 $v_{1} \dots v_{n}$ 下为对角矩阵等价于

$v_{1} \dots v_{n}$ 是 $T$ 的 $n$ 个本征向量.
$\oplus_{i} E (λ_{i}, T) = V$
存在 $n$ 个一维不变子空间直和为 $V$

这个感觉也是显然的.

内积空间

内积

二元函数<x,y>:( $V, V \to F$ )满足:

正性: $< v, v >\geq 0$
定性: $< v, v >= 0 \Leftrightarrow v = 0$
第二个位置的线性: $< u, v >$ 关于 $v$ 是线性的
共轭对称性: $< u, v >= \overset{―}{< v, u >}$

todo [think] 定性

内积 $< u, v >$ 关于 $u$ 也是线性的.
$< u, 0 >=< 0, u >= 0$

第一条用共轭对称性换到后面再换回来:

\begin{matrix} a < u_{1}, v > + b < u_{2}, v >= a \overset{―}{< v, u_{1} >} + b \overset{―}{< v, u_{2} >} \\ = \overset{―}{< v, a u_{1} + b u_{2} >} \\ =< a u_{1} + b u_{2}, v > \end{matrix}

第二条考虑线性映射 $0$ 映到 $0$ .

范数

$| | v | | =< v, v >$ 定义为向量的范数.

正交

$u ⊥ v \Leftrightarrow< u, v >= 0$

正交分解

$\forall u, v, v = \frac{u}{| | u | |^{2}} < u, v > + (v - \frac{u}{| | u | |^{2}} < u, v >)$

$< u, v >< | | u | | | | v | |$

\begin{matrix} let w = \frac{u}{| | u | |^{2}} < u, v > \\ v = w + (v - w), w ⊥ v - w \\ \Rightarrow v^{2} = w^{2} + (v - w)^{2} \leq w^{2} = \frac{< u, v >^{2}}{| | u | |^{2}} \end{matrix}

正交基

正交基,规范正交基.

正交基是两两正交的基.规范正交基就是两两正交且范数均为 $1$ 的基.

格拉姆施密特过程

任意给定一组基 $u_{1} \dots u_{n}$ ,可以构造规范正交基 $e_{1} \dots e_{n}$ .

\begin{matrix} v_{i} = u_{i} - \sum_{j = 1}^{i - 1} < u_{i}, e_{j} > e_{j} \\ e_{i} = \frac{v_{i}}{| | v_{i} | |} \end{matrix}

其实构造是很好想的,就是对于前 $i - 1$ 个正交基的空间,把第 $i$ 个去掉所有和某个基方向相同的分量,剩下的就是新的正交方向.

若 $T$ 关于 $V$ 上一组基 $v_{1} \dots v_{n}$ 由上三角矩阵,则 $T$ 关于 $V$ 上一组规范正交基有上三角矩阵.

任意复向量空间上算子关于某个规范正交基有上三角矩阵.

考虑刚才的构建过程里,每个 $span (u_{1} \dots u_{i})$ 都没有改变,所以是显然的.

而第二条可以由复向量空间上算子关于某基有上三角矩阵和第一条显然推出.

单列第二条是因为它叫舒尔定理.

里斯表示定理

对任意线性泛函 $f$ 存在 $u$ 使得 $f v =< u, v >$

设 $e_{1} \dots e_{n}$ 是一组规范正交基,则

\begin{matrix} f v = f \sum_{i = 1}^{n} < v, e_{i} > e_{i} \\ = \sum_{i = 1}^{n} < v, e_{i} > f e_{i} \\ = \sum_{i = 1}^{n} < v, f e_{i} \cdot e_{i} > \\ =< v, \sum_{i = 1}^{n} e_{i} f e_{i} > \end{matrix}

正交补

$U^{⊥} = {v | < u, v >= 0, u \in U, v \in V}$

和 $U$ 中向量正交的向量们.

$U \oplus U^{⊥} = V$
$(U^{⊥})^{⊥} = U$

取 $U$ 的一组规范正交基 $u_{1} \dots u_{n}$ ,扩充到 $V$ 的一组规范正交基 $u_{1} \dots u_{n}, v_{1} \dots v_{m}$ .

则容易发现 $U^{⊥} = span (v_{1} \dots v_{m})$ .

然后第一条是显然的.第二条的话你把 $U^{⊥}$ 的基扩充到 $V$ 的时候扩充 $u_{1} \dots u_{n}$ 就也是显然的.

正交投影

\begin{matrix} let u = w_{1} + w_{2}, w_{1} \in U, w_{2} \in U^{⊥} \\ \Rightarrow P_{U} = w_{1} \end{matrix}

即干掉垂直分量,投影到 $U$ 所在超平面上.

$P_{U}$ 是线性变换.
对 $U$ 的一组规范正交基 $e_{1} \dots e_{m}$ ,有 $P_{U} v = \sum_{i = 1}^{m} < v, e_{i} > e_{i}$

以及一些很显然的性质.

第二条看起来很显然.那么有了第二条第一条也很显然.

自伴算子,正规算子

伴随

对于算子 $T$ ,若 $\forall u, v$ , $< T u, v >=< u, T^{*} v >$ ,则 $T^{*}$ 是 $T$ 的伴随.

$(S + T)^{*} = S^{*} + T^{*}$
$(λ T)^{*} = λ T^{*}$
$(S T)^{*} = T^{*} S^{*}$
$T^*^*=T$
$null T * = range T)^{⊥}$

前三个用定义带进去即可.

第四个, $< u, T * v >=< T u, v >= \overset{―}{< T * v, u >} = \overset{―}{< v, T u >} =< T u, v >$

第五个,考虑是右边对任意 $u$ , $< T u, w >= 0$ 的所有 $w$ , $< T u, w >=< u, T * w >$ ,故 $w \in null T *$

\begin{matrix} M (T^{*}, v_{1} \dots v_{m}, u_{1} \dots u_{n}) = \overset{―}{M (T, u_{1} \dots u_{n}, v_{1} \dots v_{m})^{T}} \\ (v_{1} \dots v_{m}), (u_{1} \dots u_{n}) are regular orthogonal bases \end{matrix}

右边说的是转置再把每一项共轭.

\begin{matrix} < T u, v >=< u, T^{*} v > \\ T u = \sum_{i = 1}^{m} \sum_{j = 1}^{n} A_{j, i} < u, u_{i} > v_{j} \\ < T u, v > \\ =< \sum_{i = 1}^{m} \sum_{j = 1}^{n} A_{j, i} < u, u_{i} > v_{j}, v > \\ = \sum_{i = 1}^{m} \sum_{j = 1}^{n} A_{j, i} < u, u_{i} >< v_{j}, v > \\ =< u, \sum_{i = 1}^{m} \sum_{j = 1}^{n} A_{j, i} < v_{j}, v > u_{i} > \\ =< u, T v > \end{matrix}

其实就是直接用规范正交基写开直接做.

自伴算子

$T = T^{*}$

所以这个也是我们实对称矩阵啊.

\begin{matrix} T = T^{*} \Rightarrow {\begin{cases} T ’s eigenvalues \in R \\ \forall v, < v, T v >\in R \\ <> \end{cases} \end{matrix}

所以说书说伴随类比共轭,自伴算子类比实数啊.

English Speech

Posted on 2025-09-16 Edited on 2025-09-28

English Speech

Class 1 essay

Generative AI Enhances Human Creativity

To begin with my argument, I’ll tell a story of my own. In the summer vacation, with generative AI I developed a online notebook application whose design may be innovative and unique in my view. To complete this work I, a high school student, used AI to generative tons of code. People’s creativity comes from their whims, and most of them just evaporate for their limited capbility. However, when AI’s knowledge and skills shelter the whims from the complexity of the reality, our whims could transform into a true, creativity, rather than fade out in our mind. This is the first point.

Secondly, AI proivdes a more accessible platform for every one to gain knowledge. We can’t deny that creativity is based on knowledge. Primitive human can’t imagine the Internet while the under-educated people can’t understand generative AI’s structure as well. Today ,those difficult conception could be explained by AI, those hidden tips could be unearthed from the corner of the Internet by AI, and those people, who consist the most part of the world and can’t accept the best education, could have the most patient teacher. So AI will definitely enhance creativity by expanding the soil it comes from.

Lastly, if we omit the content I had explained and stubbornly insist AI hinder the creativity, saying like “AI’s convenient answer prevent people from thinking” or some prediction about a dark future with AI, I will say that it’s not AI hinder our creativity. When employers replace workers with ai without considering their lives, when some people depending on ai give up thinking deeply, the faughts were always on people ourselves. It may be inappropriate structure of society or the laziness of human nature. AI is just a knife and people decide who it will attack.

As a summry, AI has enhanced and will constantly enhance human’s creativity by realizing the whims that we could already quit and popularizing the knowledge be the basical root of creativity.

Class 2

Something about Chinese Culture

Unity of Human and Nature

Heaven has its seasons,erath has its engery,material has its quality, and craftmans have their skills.

drain the pounds to catch the fish

Creative Spirit of Craftsmanship
We polish the lacquer and the lacquer also polishes us.

People-oriented thinking

Filial Piety(孝顺)

the benevolent love the people.

treat the elders of others as your own, treat the kids of others as your own

harmony without uniformity

all things carreid Yin and embrace Yang,through the collection(?) of Qi to achieve harmony

gentele man seeks harmony but not uniformity

Linea Algebra Homework - Class 1

Posted on 2025-09-15 Edited on 2025-09-17

Linear Algebra Class1 Homework

T2

Going around a triangle from $(0, 0)$ to $(5, 0)$ to $(0, 12)$ to $(0, 0)$ , what are those three vectors $u, v, w$ ? What is $u + v + w$ ? What are their lengths $∥ u ∥$ , $∥ v ∥$ , and $∥ w ∥$ ? The length squared of a vector $u = (u_{1}, u_{2})$ is $∥ u ∥^{2} = u_{1}^{2} + u_{2}^{2}$ .

\begin{matrix} {\begin{cases} u = [\begin{array}{c} 5 \\ 0 \end{array}] \\ v = [\begin{array}{c} - 5 \\ 12 \end{array}] \\ w = [\begin{array}{c} 0 \\ - 12 \end{array}] \end{cases} \\ u + v + w = 0 \\ | | u | | = 5 \\ | | v | | = 13 \\ | | w | | = 12 \end{matrix}

T3

Describe geometrically (line, plane, or all of $R^{3}$ ) all linear combinations of

(a) $[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$ and $[\begin{matrix} 3 \\ 6 \\ 9 \end{matrix}]$
(b) $[\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$ and $[\begin{matrix} 0 \\ 2 \\ 3 \end{matrix}]$
© $[\begin{matrix} 2 \\ 0 \\ 0 \end{matrix}]$ and $[\begin{matrix} 0 \\ 2 \\ 2 \end{matrix}]$ and $[\begin{matrix} 2 \\ 2 \\ 3 \end{matrix}]$

(a):

a line with direction vector $[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

(b):

a plane through the origin

©:

all the 3d space

T5

If $v + w = [\begin{matrix} 5 \\ 1 \end{matrix}]$ and $v - w = [\begin{matrix} 1 \\ 5 \end{matrix}]$ , compute and draw the vectors $v$ and $w$ .

\begin{matrix} v = [\begin{matrix} 3 \\ 3 \end{matrix}] \\ w = [\begin{matrix} 2 \\ - 2 \end{matrix}] \end{matrix}

T7

Compute $u + v + w$ and $2 u + 2 v + w$ . How do you know $u, v, w$ lie in a plane?

u = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}] v = [\begin{matrix} - 3 \\ 1 \\ - 2 \end{matrix}] w = [\begin{matrix} 2 \\ - 3 \\ - 1 \end{matrix}]

These lie in a plane because $w = c u + d v$ . Find $c$ and $d$ .

\begin{matrix} u + v + w = 0 \\ 2 u + 2 v + w = [\begin{matrix} - 2 \\ - 3 \\ - 1 \end{matrix}] \\ {\begin{cases} c = - 1 \\ d = - 1 \end{cases} \end{matrix}

T11

If three corners of a parallelogram are $(1, 1)$ , $(4, 2)$ , and $(1, 3)$ , what are all three of the possible fourth corners? Draw those three parallelograms.

$(4, 0), (4, 4), (- 2, 2)$

a triangle with another three outer vertices

T13

Review Question. In xyz space, where is the plane of all linear combinations of $i = (1, 0, 0)$ and $i + j = (1, 1, 0)$ ?

equal to the linear combination of i and j( $j = (0, 1, 0)$ ).

so it’s the plane xOy

T17

What combination $c [\begin{matrix} 1 \\ 2 \end{matrix}] + d [\begin{matrix} 3 \\ 1 \end{matrix}]$ produces $[\begin{matrix} 14 \\ 8 \end{matrix}]$ ? Express this question as two equations for the coefficients c and d in the linear combination.

{\begin{cases} 14 = 1 c + 3 d \\ 8 = 2 c + 1 d \end{cases} \Rightarrow {\begin{cases} c = 2 \\ d = 4 \end{cases}

T19

Restricted only by $c \geq 0$ and $d \geq 0$ draw the “cone” of all combinations $c u + d v$ .

a angle that has the origin as the vertex(we should paint all the points between u and v black)

T23

If you look at all combinations of those $u, v, w$ , is there any vector that can’t be produced from $c u + d v + e w$ ? Different answer if $u, v, w$ are all in ______.

(1):no.

(2):a plane

Challange Problems

T24

How many corners $(\pm 1, \pm 1, \pm 1, \pm 1)$ does a cube of side 2 have in 4 dimensions? What is its volume? How many 3D faces? How many edges? Find one edge.

(1):every component of every vertex’s coordinate has two value, so there are $2^{4} = 16$ vertices.

(2):volume: $2^{4} = 16$

(3):every 3d faces means a restriction of one component, so $4 * 2 = 8$

(4):every edge means a restriction of three components, and the last component’s two value represents two vertex it connects. so $(\binom{4}{3}) 2^{3} = 32$

T25

Find two different combinations of the three vectors $u = (1, 3)$ and $v = (2, 7)$ and $w = (1, 5)$ that produce $b = (0, 1)$ . Slightly delicate question: If I take any three vectors $u, v, w$ in the plane, will there always be two different combinations that produce $b = (0, 1)$ ?

(1)

$w = - 3 u + 2 v, b = - 2 u + 2 v$ .

so for each $t$

{\begin{cases} u = - 2 - 3 t \\ v = 2 + 2 t \\ w = - t \end{cases}

is a solution

(2)

No. eg: $u = v = w = (1, 0)$

Yes when there are two vectors in ${u, v, w}$ that are linearly dependent.

T26

The linear combinations of $v = (a, b)$ and $w = (c, d)$ fill the plane unless ________. Find four vectors $u, v, w, z$ with four nonzero components each so that their combinations $c u + d v + e w + f z$ produce all vectors in four-dimensional space.

(1)

$\exists k s . t . w = k v or u = 0 or v = 0$

(2)

{\begin{cases} u = [1, 1, 1, 2] \\ v = [1, 1, 2, 1] \\ w = [1, 2, 1, 1] \\ z = [1, 1, 1, 1] \end{cases}

(do some elementary row operation on Identity Matrix)

T27

Write down three equations for $c, d, e$ so that $c u + d v + e w = b$ . Write this also as a matrix equation $A x = b$ . Can you somehow find $c, d, e$ for this $b$ ?
$u = [\begin{matrix} 2 \\ - 1 \\ 0 \end{matrix}] v = [\begin{matrix} - 1 \\ 2 \\ - 1 \end{matrix}] w = [\begin{matrix} 0 \\ - 1 \\ 2 \end{matrix}] b = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}]$

{\begin{cases} c = \frac{3}{4} \\ d = \frac{1}{2} \\ e = \frac{1}{4} \end{cases}

Math Analysis Homework - Week 1

Posted on 2025-09-15 Edited on 2025-09-28

Math Analysis Homework - Week 1

Class 1 Homework

T1

$a_{n} \leq b_{n} \leq c_{n}, lim_{n \to \infty} (c_{n} - a_{n}) = 0 \Rightarrow a_{n} 收敛$

Obviously wrong.

\begin{matrix} a_{n} = b_{n} = c_{n} = n \end{matrix}

T2

\begin{matrix} a_{n} \leq b_{n} \leq c_{n}, b_{n} 收敛, lim_{n \to \infty} (c_{n} - a_{n}) = 0 \Rightarrow a_{n} 收敛 \end{matrix}

\begin{matrix} \forall ϵ_{1} > 0, \exists N_{1} s . t . n > N_{1} \Rightarrow c_{n} - a_{n} < ϵ_{1} \\ ∴ b_{n} - a_{n} \leq c_{n} - a_{n} < ϵ_{1} \\ 又 ∵ \forall ϵ_{2} > 0 \exists N_{2} s . t . n > N_{2} \Rightarrow b_{n} - B < ϵ_{2} \\ ∴ | a_{n} - B | \leq | b_{n} - a_{n} | + | b_{n} - B | \leq ϵ + ϵ_{2} \\ ∴ ϵ_{1}, ϵ_{2} := \frac{ϵ}{2} 有 \\ \forall ϵ, N := max (N_{1}, N_{2}), n > N \Rightarrow | a_{n} - B | < ϵ \\ Q.E.D \end{matrix}

T3

\begin{matrix} lim a_{n} = A, a_{n} \neq 0 \Rightarrow lim \frac{a_{n + 1}}{a_{n}} = 1 \end{matrix}

Wrong

$a_{n} = 2^{- n}$

T4

\begin{matrix} lim_{n \to \infty} a_{n} b_{n} = 0 \Rightarrow (lim_{n \to \infty} a_{n}) (lim_{n \to \infty} b_{n}) = 0 \end{matrix}

Wrong

\begin{matrix} a_{n} = (n mod 2) \\ b_{n} = ((n + 1) mod 2) \end{matrix}

T5

\begin{matrix} lim_{n \to \infty} \frac{b_{n}}{a_{n}} = 1, lim_{n \to \infty} a_{n} = A \Rightarrow lim_{n \to \infty} b_{n} = A \end{matrix}

不妨设 $A > 0$ ,又因为取 $ϵ < A$ 可以让 $n > N$ 时 $a_{n} > 0$ ,故不妨设 $a_{n} > 0$

又 $, ϵ_{1} < 1, ϵ_{2} < A$

\begin{matrix} \forall ϵ_{1} \exists N_{1} s . t . | \frac{a_{n}}{b_{n}} - 1 | < ϵ_{1} \\ \forall ϵ_{2} \exists N_{2} s . t . | a_{n} - A | < ϵ_{2} \\ ∴ \frac{a_{n}}{b_{n}} \in (1 - ϵ_{1}, 1 + ϵ_{1}) \\ a_{n} \in (A - ϵ_{2}, A + ϵ_{2}) \\ ∴ b_{n} = \frac{a_{n}}{\frac{a_{n}}{b_{n}}} \in (\frac{A - ϵ_{2}}{1 + ϵ_{1}}, \frac{A + ϵ_{2}}{1 - ϵ_{1}}) \\ ∴ b_{n} - A \in (\frac{- A ϵ_{1} - ϵ_{2}}{1 + ϵ_{1}}, \frac{A ϵ_{1} + ϵ_{2}}{1 - ϵ_{1}}), \\ | b_{n} - A | \leq \frac{A ϵ_{1} + ϵ_{2}}{1 - ϵ_{1}} < ϵ \\ ϵ_{1} := \frac{ϵ}{100 A}, ϵ_{2} := \frac{ϵ}{100} \\ \Rightarrow | b_{n} - A | \leq \frac{A ϵ_{1} + ϵ_{2}}{1 - ϵ_{1}} = \frac{\frac{ϵ}{50}}{1 - \frac{ϵ}{100 A}} \\ let ϵ < A \Rightarrow 1 - \frac{ϵ}{100 A} > \frac{1}{50} \\ ∴ N = max (N_{1}, N_{2}) \Rightarrow n > A \Rightarrow | b_{n} - A | < ϵ \\ Q.E.D \end{matrix}

T6

\begin{matrix} lim_{n \to \infty} \frac{3 n^{2} + n}{2 n^{2} - 1} = \frac{3}{2} \end{matrix}

\begin{matrix} \frac{3 n^{2} + n}{2 n^{2} - 1} - \frac{3}{2} > \frac{3 n^{2}}{2 n^{2}} - \frac{3}{2} = 0 \\ \frac{3 n^{2} + n}{2 n^{2} - 1} - \frac{3}{2} = \frac{3 n^{2} + n - \frac{3}{2} (2 n^{2} - 1)}{2 n^{2} - 1} = \frac{n + \frac{3}{2}}{2 n^{2} - 1} \overset{n > 1}{<} \frac{2 n}{n^{2}} = \frac{2}{n} < ϵ \\ ∴ N := \frac{2}{ϵ} + 114514 \\ Q.E.D \end{matrix}

T7

\begin{matrix} lim_{n \to \infty} \sqrt{n^{2} + n} - n = \frac{1}{2} \end{matrix}

\begin{matrix} \sqrt{n^{2} + n} - n = \frac{(\sqrt{n^{2} + n} - n) (\sqrt{n^{2} + n} + n)}{\sqrt{n^{2} + n} + n} \\ = \frac{n}{\sqrt{n^{2} + n} + n} \\ = \frac{1}{1 + \sqrt{1 + \frac{1}{n}}} \\ | \frac{1}{1 + \sqrt{1 + \frac{1}{n}}} - \frac{1}{2} | \\ = \frac{1}{2} - \frac{1}{1 + \sqrt{1 + \frac{1}{n}}} \\ = \frac{\sqrt{1 + \frac{1}{n}} - 1}{2 (1 + \sqrt{1 + \frac{1}{n}})} \\ < \sqrt{1 + \frac{1}{n}} - 1 \\ \overset{Bernoulli Inequality}{<} 1 + \frac{1}{2 n} - 1 \\ = \frac{1}{2 n} \\ ∴ N := \frac{1}{2 ϵ} + 100 \\ Q.E.D \end{matrix}

T8

\begin{matrix} \frac{n^{2} \arctan (n)}{1 + n^{2}} = \frac{π}{2} \end{matrix}

\begin{matrix} \frac{n^{2} \arctan (n)}{1 + n^{2}} \\ = \frac{n^{2}}{1 + n^{2}} \arctan (n) \end{matrix}

\begin{matrix} lim_{n \to \infty} \frac{n^{2}}{1 + n^{2}} = 1 \end{matrix}

\begin{matrix} lim | \frac{n^{2}}{1 + n^{2}} - 1 | = \frac{1}{1 + n^{2}} < \frac{1}{n} \\ N := \frac{1}{ϵ} + 100 \\ Q.E.D \end{matrix}

\begin{matrix} lim_{n \to \infty} \arctan (n) = \frac{π}{2} \end{matrix}

\begin{matrix} N := \tan (\frac{π}{2} - \frac{ϵ}{2}) \Rightarrow \\ \forall ϵ, | \arctan (n) - \frac{π}{2} | = \frac{π}{2} - \arctan (n) = \frac{ϵ}{2} < ϵ \\ Q.E.D \end{matrix}

\begin{matrix} (lim a_{n}) (lim_{n \to \infty} b_{n}) = X, a_{n} > 0, b_{n} > 0 \Rightarrow lim_{n \to \infty} a_{n} b_{n} = X \end{matrix}

\begin{matrix} A := lim a_{n}, B := lim b_{n} \\ \forall ϵ_{1}, \exists N_{1} s . t . n > N_{1} \Rightarrow | a_{n} - A | < ϵ \\ \forall ϵ_{2}, \exists N_{2} s . t . n > N_{2} \Rightarrow | b_{n} - B | < ϵ \\ ∴ n > max (N_{1}, N_{2}) \Rightarrow \\ a_{n} \in (A - ϵ_{1}, A + ϵ_{1}), b_{n} \in (B - ϵ_{2}, B + ϵ_{2}) \\ \Rightarrow \\ a_{n} b_{n} \in ((A - ϵ_{1}) (B - ϵ_{2}), (A + ϵ_{1}) (B + ϵ_{2})) \\ | a_{n} b_{n} - A B | < A ϵ_{2} + B ϵ_{1} + ϵ_{1} ϵ_{2} \\ \forall ϵ, ϵ_{2} := \frac{ϵ}{4 A}, ϵ_{1} := \frac{ϵ}{4 B} \\ ∴ | a_{n} b_{n} - A B | = \frac{ϵ}{2} + \frac{ϵ^{2}}{16 A B} \overset{ϵ < A B}{<} ϵ \\ Q.E.D \end{matrix}

\begin{matrix} lim_{n \to \infty} \frac{n^{2} \arctan (n)}{1 + n^{2}} = lim_{n \to \infty} \frac{n^{2}}{1 + n^{2}} lim_{n \to \infty} \arctan (n) = 1 \times \frac{π}{2} = \frac{π}{2} \\ Q.E.D \end{matrix}

Class 2

T1

$lim_{n \to \infty} \frac{(- 2)^{n} + 3^{n}}{(- 2)^{n + 1} + 3^{n + 1}}$

lim_{n \to \infty} \frac{(- 2)^{n} + 3^{n}}{(- 2)^{n + 1} + 3^{n + 1}} lim_{n \to \infty} = \frac{1}{3} \frac{(\frac{- 2}{3})^{n} + 1}{(\frac{- 2}{3})^{n + 1} + 1} lim_{n \to \infty} = \frac{1}{3}

T2

$lim_{n \to \infty} [\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n (n + 1)}]$

= lim_{n \to \infty} \sum_{i = 1}^{n} \frac{1}{i (i + 1)} = lim_{n \to \infty} \sum_{i = 1}^{n} \frac{1}{i} - \frac{1}{i + 1} = lim_{n \to \infty} 1 - \frac{1}{n + 1} = 1

T3

$lim_{n \to \infty} [\frac{1}{\sqrt{n^{2} + 1}} + \frac{1}{\sqrt{n^{2} + 2}} + \dots + \frac{1}{\sqrt{n^{2} + n}}]$

\begin{matrix} \sum_{i = 1}^{n} \frac{1}{\sqrt{n^{2} + 1}} \leq \sum_{i = 1}^{n} \frac{1}{\sqrt{n^{2} + i}} \leq \sum_{i = 1}^{n} \frac{1}{\sqrt{n^{2} + n}} \\ \overset{Squeeze Theorem}{⟹} \\ lim_{n \to \infty} \sum_{i = 1}^{n} \frac{1}{\sqrt{n^{2} + 1}} \leq L = lim_{n \to \infty} \sum_{i = 1}^{n} \frac{1}{\sqrt{n^{2} + i}} \leq lim_{n \to \infty} \sum_{i = 1}^{n} \frac{1}{\sqrt{n^{2} + n}} \\ \Rightarrow lim_{n \to \infty} \frac{n}{\sqrt{n^{2} + 1}} \leq L \leq lim_{n \to \infty} \frac{n}{\sqrt{n^{2} + n}} \\ \Rightarrow lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^{2}}}} \leq L \leq lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n}}} \\ \Rightarrow L = 1 \end{matrix}

T4

$lim_{n \to \infty} \sqrt[n]{n^{2} - n + 2}$

Obviously: $n > 2 \Rightarrow n^{2} - n + 2 > 8 > 1$

\begin{matrix} | (n^{2} - n + 2)^{\frac{1}{n}} - 1 | = (n^{2} - n + 2)^{\frac{1}{n}} - 1 < ϵ \\ \Leftarrow n^{2} - n + 2 < (1 + ϵ)^{n} \\ \Leftarrow F (x) = n^{2} - n + 2 < 1 + n ϵ + \frac{(n^{2} - n) ϵ^{2}}{2} + \frac{n (n - 1) (n - 2)}{6} ϵ^{3} = G (x) \\ \exists C_{1} (ϵ), C_{2} (ϵ) \\ s . t . \\ F (x) < C_{1} (ϵ) n^{2}, G (x) > C_{2} (ϵ) n^{3} \\ ∴ N := \frac{C_{1} (ϵ)}{C_{2} (ϵ)} + 1 \Rightarrow (n > N \Rightarrow (n^{2} - n + 2)^{\frac{1}{n}} - 1 < ϵ) \\ ∴ lim_{n \to \infty} \sqrt[n]{n^{2} - n + 2} = 1 \end{matrix}

T5

lim_{n \to \infty} \sqrt[n]{\arctan (n)}

\begin{matrix} 1 < \arctan (n) < \frac{π}{2} \\ \Rightarrow 1 < \sqrt[n]{\arctan (n)} < \sqrt[n]{\frac{π}{2}} \\ \begin{array}{ll} 1 < \sqrt[n]{\arctan (n)} < \sqrt[n]{\frac{π}{2}} \\ lim_{n \to \infty} 1 = lim_{n \to \infty} \sqrt[n]{\frac{π}{2}} \end{array}} \overset{Squeeze Theorem}{⟹} lim_{n \to \infty} \sqrt[n]{\arctan (n)} = 1 \end{matrix}

T6

lim_{n \to \infty} \sqrt[n]{2 \sin^{2} (n) + \cos^{2} (n)}

同上一题,里面有界,是 $1$

T7

$lim_{n \to \infty} \frac{[n a_{n}]}{n}$ , 这里 $lim_{n \to \infty} a_{n} = a$ .

\begin{array}{ll} \frac{n a_{n} - 1}{n} < \frac{[n a_{n}]}{n} < \frac{n a_{n} + 1}{n} \\ lim_{n \to \infty} \frac{n a_{n} + 1}{n} = lim_{n \to \infty} a_{n} + lim_{n \to \infty} \frac{1}{n} = a \\ lim_{n \to \infty} \frac{n a_{n} - 1}{n} = lim_{n \to \infty} a_{n} - lim_{n \to \infty} \frac{1}{n} = a \end{array}} \overset{Squeeze Theorem}{⟹} lim_{n \to \infty} \frac{[n a_{n}]}{n} = a_{n}

T8

证明 $a_{n} = \frac{2 n + (- 1)^{n} n}{3 n + 1}$ 发散

\begin{matrix} \begin{array}{ll} lim_{n \to \infty} a_{2 n} = lim_{n \to \infty} \frac{6 n}{6 n + 1} = 1 \\ lim_{n \to \infty} a_{2 n + 1} = lim_{n \to \infty} \frac{2 n + 1}{6 n + 4} = \frac{1}{3} \neq 1 \end{array}} \\ \Rightarrow {a_{n}} 发散 \end{matrix}

T9

a_{n} \neq 0, \frac{a_{n + 1}}{a_{n}} > 0, lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 0 \Rightarrow \exists N s . t . n > N \Rightarrow {a_{n}} 单调

\begin{matrix} use ϵ_{1} = 1, \exists N_{1} \\ s . t . \\ lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} < 1 \\ exact a_{n + 1} < a_{n} \end{matrix}

T10

设 $lim_{n \to \infty} (x_{n} - x_{n - 1}) = d$ , 证明: $lim_{n \to \infty} \frac{x_{n}}{n} = d$ .

\begin{matrix} \forall ϵ_{1}, \exists N_{1} s . t . n > N_{1} \Rightarrow | x_{n} - x_{n - 1} - d | < ϵ_{1} \\ \Leftrightarrow x_{n} - x_{n - 1} \in [d - ϵ_{1}, d + ϵ_{1}] \\ ∴ x_{n} = x_{N_{1}} + \sum_{i = N_{1} + 1}^{n} (x_{i} - x_{i - 1}) \\ \in [x_{N_{1}} + (n - N_{1}) (d - ϵ_{1}), x_{N_{1}} + (n - N_{1}) (d + ϵ_{1})] \\ ∴ \frac{x_{n}}{n} \in [\frac{x_{N_{1}} - N_{1} (d - ϵ_{1})}{n} + d - ϵ_{1}, \frac{x_{N_{1}} - N_{1} (d + ϵ_{1})}{n} + d + ϵ_{1}] \\ \forall ϵ, \frac{x_{n}}{n} - d \in [\frac{x_{N_{1}} - N_{1} (d - ϵ_{1})}{n} - ϵ_{1}, \frac{x_{N_{1}} - N_{1} (d + ϵ_{1})}{n} + ϵ_{1}] \\ ϵ_{1} := \frac{ϵ}{2}, n := \frac{2 x_{N_{1}} - N_{1} (d - ϵ_{1})}{ϵ} \\ \Rightarrow | \frac{x_{n}}{n} - d | < ϵ \end{matrix}

T11

设 $lim_{n \to \infty} a_{n} = a (a > 0, n \in N)$ , 证明: $lim_{n \to \infty} \sqrt[n]{a_{1} a_{2} \dots a_{n}} = a$ , 并由此证明:

若 $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = a (a > 0, n \in N)$ ,则 $lim_{n \to \infty} \sqrt[n]{a_{n}} = a$ ;
$lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n} = \frac{1}{e}$

(1)

Solution 1

\begin{matrix} | \sqrt[n]{\prod_{i} a_{i}} - a | < ϵ \\ \Leftrightarrow a - ϵ < \sqrt[n]{\prod_{i} a_{i}} < a + ϵ \\ \Leftrightarrow (a - ϵ)^{n} < \prod_{i} a_{i} < (a + ϵ)^{n} \\ \forall ϵ_{1}, \exists N_{1} s . t . n > N_{1} \Rightarrow a_{n} \in [a - ϵ_{1}, a + ϵ_{1}] \\ \prod_{i} a_{i} = (\prod_{i = 1}^{N_{1}} a_{i}) (\prod_{i = N_{1} + 1}^{n} a_{i}) \in [A (a - ϵ_{1})^{n - N_{1}}, A (a + ϵ_{1})^{n - N_{1}}] \\ let ϵ_{1} := \frac{ϵ}{2}, Consider A (a - ϵ_{1})^{n - N_{1}} > (a - ϵ)^{n} : \\ \Leftrightarrow \frac{A}{(a - ϵ_{1})^{N_{1}}} > (\frac{a - ϵ}{a - ϵ_{1}})^{n} \\ Since \frac{a - ϵ}{a - ϵ_{1}} < 1, \exists N_{2} s . t . n > N_{2} \Rightarrow 不等式成立 \\ 右侧同理有 N_{3} \\ ∴ N := N_{1} + N_{2} + N_{3} \\ s . t . \\ n > N \\ \Rightarrow (a - ϵ)^{n} < A (a - ϵ_{1})^{n - N_{1}} < \prod_{i} a_{i} < A (a + ϵ_{1})^{n - N_{1}} < (a + ϵ)^{n} \\ \Rightarrow | \sqrt[n]{\prod_{i} a_{i}} - a | < ϵ \\ Q.E.D \end{matrix}

Solution 2

唐. 可以用调和均值/算数均值夹两边.

(2)

令 $b_{n} = \frac{a_{n + 1}}{a_{n}}$ ,则问题转化为(1).

(3)

\begin{matrix} e = lim_{n \to \infty} (1 + \frac{1}{n})^{n} \\ \Rightarrow \frac{1}{e} = lim_{n \to \infty} (\frac{n}{n + 1})^{n} \\ a_{n} := (\frac{n}{n + 1})^{n} 检验符合引理 \\ Q.E.D \end{matrix}

T12

设 $lim_{n \to \infty} x_{n} = + \infty$ , 证明: $lim_{n \to \infty} \frac{x_{1} + x_{2} + \dots + x_{n}}{n} = + \infty$ .

\begin{matrix} \forall X, X_{1} := X + 1 \\ lim_{n \to \infty} x_{n} = \infty \Rightarrow \exists N_{1} s . t . n > N_{1} \Rightarrow x_{n} > X_{1} = X + 1 \\ for n > N_{1}, \frac{\sum_{i = 1}^{n} x_{i}}{n} = \frac{\sum_{i = 1}^{N_{1}} x_{i} + \sum_{i = N_{1} + 1}^{n} X + 1}{n} \\ > (1 - \frac{N_{1}}{n}) (X + 1) \\ n := N_{1} (X + 1) + 100 \\ \Rightarrow \frac{\sum_{i = 1}^{n} x_{i}}{n} > X \end{matrix}

T13

a_{n} > 0, lim_{n \to \infty} \frac{a_{n}}{a_{n + 1} + a_{n + 2}} = 0 \Rightarrow a_{n} is unbounded

反证,设 $\exists M$ 令 $0 < a_{n} < M$ .

\begin{matrix} ϵ = \frac{1}{5}, \frac{a_{n}}{a_{n + 1} + a_{n + 2}} < \frac{1}{5} \Rightarrow a_{n + 1} + a_{n + 2} > 5 a_{n} \\ ∴ max ({a_{n + 1}, a_{n + 2}}) > 2 a_{n} \\ ∴ b_{1} = N + 1, b_{i} = a_{b_{i - 1} + 1}, a_{b_{i - 1} + 2} 中较大的一个的下标 \Rightarrow a_{b_{i}} > 2^{i} a_{b_{1}} \end{matrix}

故 $a$ 有发散子列, $a$ 发散.

T14

设数列 ${x_{n}}$ 单调增加, $lim_{n \to \infty} \frac{x_{1} + x_{2} + \dots + x_{n}}{n} = a$ , 证明: $lim_{n \to \infty} x_{n} = a$ .

若存在 $x_{N} > a$ ,则 $n > N$ 时 $x_{n} > a$ , $lim_{n \to \infty} x_{N} > a$ ,则 $\exists A \in (x_{N}, A)$ . 并有 $\forall i \geq N, x_{i} > A > a$ ,则

\begin{matrix} lim_{n \to \infty} \frac{\sum_{i = 1}^{n} x_{i}}{n} \\ > lim_{n \to \infty} \frac{\sum_{i = 1}^{N} a_{i}}{n} + \frac{n - N}{n} (a + (A - a)) = A > a \end{matrix}

矛盾,故 $\forall i, x_{i} < a$

于是

\begin{array}{ll} \frac{\sum_{i = 1}^{n} x_{i}}{n} < x_{n} < a \\ lim_{n \to \infty} a = a \\ lim_{n \to \infty} \frac{\sum_{i = 1}^{n} x_{i}}{n} = a \end{array}} \Rightarrow lim_{n \to \infty} x_{n} = a

Class 3

T1

$lim_{n \to \infty} \frac{1 + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}}}{\ln \sqrt{n}}$

\begin{matrix} lim_{n \to \infty} \frac{1 + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}}}{\ln \sqrt{n}} \\ \overset{Stolz Theorem}{\Leftarrow} lim_{n \to \infty} \frac{\frac{2}{\sqrt{n}}}{\ln \frac{n}{n - 1}} \\ \overset{x > 1 \Rightarrow \ln (x) > 2 \frac{x - 1}{x + 1}}{>} lim_{n \to \infty} \frac{\frac{2}{\sqrt{n}}}{\frac{2}{2 n - 1}} \\ = + \infty \end{matrix}

T2

$lim_{n \to \infty} \frac{1 + \sqrt{2} + \sqrt[3]{3} + \dots + \sqrt[n]{n}}{n}$

\begin{matrix} lim_{n \to \infty} \frac{1 + \sqrt{2} + \sqrt[3]{3} + \dots + \sqrt[n]{n}}{n} \\ \overset{Stolz Theorem}{\Leftarrow} lim_{n \to \infty} \frac{\sqrt[n]{n}}{1} \\ = 1 \end{matrix}

T3

$lim_{n \to \infty} \frac{a_{1} + 2 a_{2} + \dots + n a_{n}}{\sum_{i = 1}^{n} i}$ (已知 $lim_{n \to \infty} a_{n} = a$ )

\begin{matrix} lim_{n \to \infty} \frac{a_{1} + 2 a_{2} + \dots + n a_{n}}{\sum_{i = 1}^{n} i} \\ \overset{Stolz Theorem}{\Leftarrow} lim_{n \to \infty} \frac{n a_{n}}{n} \\ = a \end{matrix}

T4

计算极限 $lim_{n \to \infty} (n!)^{\frac{1}{n^{2}}}$ .

\begin{matrix} \begin{array}{ll} 1 < (n!)^{\frac{1}{n^{2}}} < (n^{n})^{\frac{1}{n^{2}}} = n^{\frac{1}{n}} \\ lim_{n \to \infty} 1 = lim_{n \to \infty} n^{\frac{1}{n}} = 1 \end{array}} \\ \overset{Squeeze Theorem}{⟹} lim_{n \to \infty} (n!)^{\frac{1}{n^{2}}} = 1 \end{matrix}

T5

设 $x_{n} = \frac{1}{n^{2}} \sum_{k = 0}^{n} \ln (\binom{n}{k})$ , $n = 1, 2, \dots$ , 求极限 $lim_{n \to \infty} x_{n}$ .

\begin{matrix} \frac{1}{n^{2}} \sum_{k = 0}^{n} \ln (\binom{n}{k}) \\ \overset{Stolz Theorem}{\Leftarrow} \frac{\sum_{i = 0}^{n} \ln (\binom{n}{i}) - \sum_{i = 0}^{n - 1} \ln (\binom{n - 1}{i})}{2 n - 1} \\ = \frac{\sum_{i = 0}^{n - 1} \ln (\frac{n}{n - i})}{2 n - 1} \\ = \frac{\ln (\frac{n^{n}}{n!})}{2 n - 1} \\ = \frac{n \ln (\frac{n}{\sqrt[n]{n!}})}{2 n - 1} \\ According to homework class-2: \\ lim_{n \to \infty} \frac{\sqrt[n]{n!}}{n} = \frac{1}{e} \\ \Rightarrow lim_{n \to \infty} \frac{n \ln (\frac{n}{\sqrt[n]{n!}})}{2 n - 1} \\ = lim_{n \to \infty} \frac{n}{2 n - 1} lim_{n \to \infty} \ln (\frac{n}{\sqrt[n]{n!}}) \\ = \frac{1}{2} \end{matrix}

还是用了连续性/kk

T6

设 $lim_{n \to \infty} n (A_{n} - A_{n - 1}) = 0$ , 试证: 当极限 $lim_{n \to \infty} \frac{A_{1} + A_{2} + \dots + A_{n}}{n}$ 存在时, $lim_{n \to \infty} A_{n} = lim_{n \to \infty} \frac{A_{1} + A_{2} + \dots + A_{n}}{n}$ .

\begin{matrix} x_{n} := \sum_{i = n}^{n} Δ A_{i} i \\ = \sum_{i = 1}^{n} i (A_{i} - A_{i - 1}) \\ = n A_{n} - \sum_{i = 1}^{n - 1} A_{i} \\ lim_{n \to \infty} \frac{x_{n}}{n} \\ \overset{Stolz Theorem}{\Leftarrow} \frac{x_{n} - x_{n - 1}}{n - (n - 1)} = n Δ A_{n} = 0 \\ ∴ lim_{n \to \infty} A_{n} - lim_{n \to \infty} \frac{\sum_{i = 1}^{n - 1} A_{i}}{n - 1} = lim_{n \to \infty} \frac{x_{n}}{n} = 0 \\ lim_{n \to \infty} A_{n} = a \end{matrix}

Linear Algebra

Posted on 2025-09-15 Edited on 2025-09-26

Linear algebra

A fun question

In n-D space we can found at most $n + 1$ vector $v_{1} \dots v_{n + 1}$ such that: $\forall i \neq j, v_{i} v_{j} < 0$

An example in 3-D space(the one on the book)

The construction is Obviously( $C H_{4}$ )

Chosen a vector, the others must be in a semisphere. We say two semisphere is seperated by plane A.

And we notice that: if two vector’s shadow on plane A construct a acute angle, their dot product must be positive, transformed it into a 2-D problems which is easy to solve.

Generalize to n-D

We choose a vector $v_{1}$ , and it can be written as $[1, 0, \dots 0]$ (with some rotation)

so $\forall v_{i}, i > 1, v_{i} v_{1} < 0 \Rightarrow v_{i, 1} < 0 \Rightarrow v_{i, 1} v_{j, 1} > 0 \Rightarrow v_{i} v_{j} - v_{i, 1} v_{j, 1} < 0$

then transformed it into the (n-1)-D situation.

The construction can be easily give during the induction

Another Proof

given by Bing!

Lemma: Radon Partition

In n-D space, $n + 2$ point could be divided to two convex hull with intersection.

$n + 2$ vector must be dependent: $\exists c_{i} s . t . \sum_{i = 1}^{n + 2} c_{i} x_{i} = 0; \sum_{i} x_{i} = 0$ (the second condition can be satisfied by add another all-1 dimension).

so divide the vector by sign of $c_{i}$ we got: $v = \sum_{i \in A} c_{i} x_{i} = \sum_{j \in B} c_{j} x_{j}$ ,so divide the eqution by $\sum_{i \in A} c_{i}$ , you get one point( $v$ ) in the intersection.

We noticed $0 < v^{2} = (\sum_{i \in A} c_{i} x_{i}) \cdot (\sum_{j \in B} c_{j} x_{j}) < 0$ , contradiction!

Operator’s Left Inverse and Right Inverse

Existance

对算子 $T$ 来说,左逆存在等价于右逆存在.

Proof 1

注意到左逆存在等价于 $T$ 是单射,右逆存在等价于 $T$ 是满射.

又因为 $T$ 单射等价于 $T$ 满射等价于 $T$ 双射,所以左逆存在等价于右逆存在且两个逆一定相同.

Proof 2

左逆推出 $T$ 分解成初等行变换矩阵,然后一个一个取逆推右逆.

Math Analysis

Posted on 2025-09-15 Edited on 2025-09-29

Math analysis

Class 1 Some Inequality and Def of limit

Triangle Inequality

a, b \in R \Rightarrow | | a | - | b | | \leq | a + b | \leq | a | + | b |

\begin{matrix} - | a | | b | \leq a b \leq | a | | b | \Rightarrow \\ {| a |}^{2} - 2 | a | | b | + {| a + b |}^{2} \leq a^{2} + 2 a b + b^{2} \leq {| a |}^{2} + 2 | a | | b | + {| a + b |}^{2} \Rightarrow \\ (| a | - | b |)^{2} \leq (a + b)^{2} \leq (| a | + | b |)^{2} \end{matrix}

向量的模长也满足也可以用这个做法.

Cauchy-Schwarz Inequality

\begin{matrix} 0 < x_{i} \in R \Rightarrow \frac{\sum_{i} x_{i}}{n} \geq \sqrt[n]{\prod_{i} x_{i}} \geq \frac{n}{\sum_{i} \frac{1}{x_{i}}} \end{matrix}

显然取 $x_{i} = \frac{1}{x_{i}}$ 能用左边不等式推右边只证左边

归纳, $2$ 的幂次显然吧

对非 $2$ 的幂次 $n$ ,用 $X = \sqrt[n]{\prod_{i} x_{i}}$ 补完,直接套

Array’s Limit

\begin{matrix} l i m_{n \to \infty} a_{n} = A \Leftrightarrow \\ \forall ϵ > 0 \exists N s . t . n \geq N \Rightarrow | a_{n} - A | < ϵ \end{matrix}

注意

$N = N (ϵ)$
$N$ 不唯一
$ϵ$ 可限制在任意 $(0, a), a > 0$

Eg 1

$lim_{n \to \infty} n^{\frac{1}{n}} = 1$

\begin{matrix} | n^{\frac{1}{n}} - 1 | \\ = | (\sqrt{n} \sqrt{n})^{\frac{1}{n}} - 1 | \\ = | (\prod_{i} 1 \cdot \sqrt{n} \sqrt{n})^{\frac{1}{n}} - 1 | \\ \leq | \frac{n - 2 + 2 \sqrt{n}}{n} - 1 | \\ \leq \frac{2}{\sqrt{n}} \end{matrix}

Or try this:

\begin{matrix} | n^{\frac{1}{n}} - 1 | = n^{\frac{1}{n}} - 1 < ϵ \\ \Leftarrow n^{\frac{1}{n}} < (ϵ + 1) \\ \Leftarrow n < (1 + ϵ)^{n} \\ \Leftarrow n < 1 + ϵ n + \frac{ϵ^{2} n (n - 1)}{2} \end{matrix}

Solve the equation, it’s a parabola with upward opening so the solution exists.

[think] At this stage, we cannot say $lim g (f (n)) = g (lim f (n))$ which depends on continuity of function. But you sometimes can rewrite the proof with inequality.

Class 2

About $Q$

\begin{matrix} \sqrt{n} \in Z \cup Q^{C} \end{matrix}

$p^{2} = q^{2} n$ Integer factorization

More natural than textbook but it doesn’t depend on Integer factorization.

$\frac{p}{q}$ is finite decimal or repeating decimal

Simulate how the division is done and consider the remainder will be repeated.

About Cardinality

A Example

Find a bijection of $[0, 1]$ and $(0, 1)$

\begin{matrix} f (x) : (0, 1) \to [0, 1] = {\begin{cases} 0, x = \frac{1}{2} \\ 1, x = \frac{1}{3} \\ \frac{1}{n - 2}, x = \frac{1}{n} \\ x, otherwise \end{cases} \end{matrix}

Wow!

Cardinality is comparable

\begin{matrix} \forall A, B, \exists f (x) : A \to B \\ s . t . \\ f is injective or surjective \end{matrix}

(承认选择公理)

首先选择公理出良序定理,找到 $A, B$ 的良序 $(A, <), (B, <)$ .

然后进行超限归纳,把 $A$ , $B$ 按照良序(保证了每个元素存在序里的后继),最小元素匹配,然后剩下的次小元素匹配,这么做下去.

然后这个看起来进行的是普通的自然数归纳显然是错的. 你需要用超限归纳也就是在序上做归纳.

然后问题来到序是什么.

序的结构是这样的首先是自然数自然数定义是 $0$ 定义为空集开始然后定义 $succ (S) = S \cup {S}$

然后定义完所有自然数后定义 $ω = ⋃_{i} i$ ,就是把所有的自然数的集合并起来(显然它包含所有的自然数). 然后我们可以接下来用后继的定于去定义 $ω + 1, ω + 2 \dots$ ,并且你又可以把它们并起来得到 $ω \times 2$ ,不断走后继, $ω, 2 ω, 3 ω \dots$ 可以变成 $ω^{2}$ ,又有 $ω^{3} \dots ω^{ω}$ 等等

基本上是每个层次的运算完了之后进下一个层次的序数构造总之它看起来包含了各种各样的无穷,可以应付所有大小的集合.

然后我们要在序数的结构上做归纳法,就要证明: $x \to succ (x)$ ,还要证明极限这一把也对,就是 $a_{1} \dots a_{n} \to \cup_{i} a_{i}$ 这个操作(也就是对某个极限序数,如果所有它以前的序数推到它自己)合法,就满足了你可以不断到下一极限.这样推出对全体元素合法. 这就是超限归纳法.

那你对照一下我们的归纳就是一一对应啊,所以是合法的. 就结束了.

然后我们刚才是说明了可以借良序去给两个集合配对,那么你就一定能找到一个对另一个的单射,所以一定可比.

Cantor-Bernstein-Schröder Theorem

$Card (A) \leq Card (B), Card (B) \leq Card (A) \Rightarrow Card (A) = Card (B)$

或者表达为

\begin{matrix} \begin{array}{ll} \forall A, B, g : A \to B, f : B \to A \\ f, g is injective \end{array}} \Rightarrow \exists h : A \leftrightarrow B is bijective \end{matrix}

考虑从任意元素 $u \in A$ 可以引出一条链: $u \to f (u) \to g (f (u) \to f (g (f (u))) \to \dots$ .
同理 $v \in B$ 开始的链 $v \to g (v) \to f (g (v)) \to g (f (g (v))) \to \dots$ .

注意到所有元素一定都在某条链上(映射).每个点度数一定至多一进一出(单射).

于是每条链上构造一个双射(显然的)拼起来即可.

$Q.E.D$

其实这些应该算集合论还是什么?

Class 3 Properties of Limit

Uniqueness:Obviously
Local boundedness: 取 $ϵ = 1$ 得 $N$ , $N$ 后面显然,前面有限项也显然.
About subsequence:
- about any subsequence
- about some subsequence whose union is the sequence

Limitaion’s Calculation

lim_{n \to \infty} a_{n} = A, lim_{n \to \infty} b_{n} = B

lim_{n \to \infty} a_{n} b_{n} = A B

\begin{matrix} | a_{n} b_{n} - A B | = \\ | a_{n} b_{n} - a_{n} B + a_{n} B - A B | \\ \leq | a_{N} | | b_{n} - B | + | B | | a_{n} - A | \end{matrix}

显然 $a_{n}, b_{n}$ 有界,令 $M$ 是他俩共同的界,取 $ϵ^{'} = \frac{1}{114514 M + 114514}$ 用到 $a_{n}, b_{n}$ 上即证.

lim_{n \to \infty} \frac{a_{n}}{b_{n}} = \frac{A}{B}

先上一个保号性干掉分母上出现 $0$ 的事.

证取倒数:

\begin{matrix} | \frac{1}{b_{n}} - \frac{1}{B} | < ϵ \\ \Leftarrow | B - b_{n} | < ϵ b_{n} B \\ \Leftarrow ϵ^{'} = \frac{ϵ}{B M + 114514} (| M | > b_{n}) \end{matrix} Q.E.D

那转化到乘法是显然的.

无穷小

基本就是收敛到 $0$ 的数列啊.

Eg

{\begin{cases} lim_{n \to \infty} a_{n} = a \\ lim_{n \to \infty} b_{n} = b \end{cases} \Rightarrow lim_{n \to \infty} \frac{\sum_{i} a_{i} b_{n - i}}{n} = a b

Solution 1

$b = 0$ 时,考虑 $a 有界 M$ .

\begin{matrix} lim_{n \to \infty} \frac{\sum_{i} a_{i} b_{n - i}}{n} \\ \leq lim_{n \to \infty} M \frac{b_{n - i}}{n} = 0 \end{matrix}

$b \neq 0$ , $b_{i}^{'} = b_{i} - b$ ,有

\begin{matrix} lim_{n \to \infty} \frac{\sum_{i} a_{i} b_{n - i}}{n} = lim_{n \to \infty} \frac{\sum_{i} a_{i} b_{n - i}^{'}}{n} + \frac{\sum_{i} a_{i}}{n} b = 0 + a b \end{matrix}

\begin{matrix}  \end{matrix}

Solution 2

不妨设 $a, b > 0$ ,则有界性可以得到后面是 $a_{n}, b_{n} > 0$

考虑影响值的肯定是中间的项,所以直接拆,取 $ϵ$ ,对 $a, b$ 可以得到 $N_{1}$

\begin{matrix} lim_{n \to \infty} \frac{\sum_{i} a_{i} b_{n - i}}{n} = \\ = lim_{n \to \infty} \frac{\sum_{i = 1}^{N_{1}} a_{i} b_{n - i} + \sum_{i = 1}^{N_{1}} b_{i} a_{n - i} + \sum_{i = N_{1} + 1}^{n - N_{1} - 1} a_{i} b_{n - i}}{n} \\ = lim_{n \to \infty} \frac{\sum_{i = 1}^{N_{1}} a_{i} b_{n - i}}{n} \\ + lim_{n \to \infty} \frac{\sum_{i = 1}^{N_{1}} b_{i} a_{n - i}}{n} \\ + lim_{n \to \infty} \frac{\sum_{i = N_{1} + 1}^{n - N_{1} - 1} a_{i} b_{n - i}}{n} \\ \in (lim_{n \to \infty} \frac{n - 2 N_{1} - 1}{n} (a - ϵ) (b - ϵ) \\ , lim_{n \to \infty} \frac{n - 2 N_{1} - 1}{n} (a + ϵ) (b + ϵ)) \\ = ((a - ϵ) (b - ϵ), (a + ϵ) (b + ϵ)) \end{matrix}

后面显然.`

Class 4

eg1

\begin{matrix} a > 1, k \in N^{*}, lim_{n \to \infty} \frac{n^{k}}{a^{n}} = 0 \end{matrix}

$a^{n} = (1 + b)^{n}$ 展开会出现 $n$ 的任意次方,然后显然.

Stolz

\begin{matrix} \begin{array}{ll} y_{n} ↑, lim_{n \to \infty} y_{n} = \infty \\ lim_{n \to \infty} \frac{x_{n} - x_{n - 1}}{y_{n} - y_{n - 1}} = a \in [- \infty, + \infty] \end{array}} \\ \Rightarrow lim_{n \to \infty} \frac{x_{n}}{y_{n}} = a \end{matrix}

先简化,不妨设 $a \geq 0, \frac{Δ x_{n}}{Δ y_{n}} > 0$

然后可以证 $a = 0$ 时,你取 $ϵ_{1}$ 得 $N_{1}$ ,变成 $\frac{Δ x_{n}}{Δ y_{n}} < ϵ$ .于是

\begin{matrix} \frac{x_{n}}{y_{n}} \\ = \frac{x_{N_{1}} + \sum_{i = N_{1} + 1}^{n} Δ x_{i}}{y_{N_{1}} + \sum_{i = N_{1} + 1}^{n} Δ y_{i}} \\ \leq \frac{x_{N_{1}} + ϵ \sum_{i = N_{1} + 1}^{n} Δ y_{i}}{y_{N_{1}} + \sum_{i = N_{1} + 1}^{n} Δ y_{i}} \\ \leq \frac{x_{N_{1}} + ϵ \sum_{i = N_{1} + 1}^{n} Δ y_{i}}{\sum_{i = N_{1} + 1}^{n} Δ y_{i}} \end{matrix}

因为你 $Y = \sum_{i = N_{1} + 1}^{n} Δ y_{i} \to + \infty$ ,所以一定可以有 $ϵ Y > x_{N_{1}}$ .就能证 $< 2 ϵ$ ,你再取一下 $ϵ$ 就得证了.

然后我说 $0 < a \in R$ 时你直接 $x^{'} = x - a y$ 就用结论, $a = + \infty$ 的时候你取倒数用结论,就做完了.

Class 5

单调有界数列有极限

\begin{matrix} \begin{array}{ll} a_{i} > a_{i - 1} \\ a_{i} < M \end{array}} \Rightarrow lim_{n \to \infty} a_{i} exists \end{matrix}

取 ${a_{i}}$ 上确界 $M_{0}$ ,则任意 $ϵ$ ,取 $M^{'} = M_{0} - ϵ$ ,由确界知 $\exists i s . t . a_{i} \in (M^{'}, M_{0}]$ ,则 $\forall n > i, | a_{n} - M | < ϵ$ .

然后书上因为没教你上确界,试图用无限小数说明.那其实就是你一位一位考虑,就先这一位增加到最大,然后进入下一位,容易发现最后区间长度趋近于 $0$

其实无限小数就是区间套啊.

e

\begin{matrix} a_{n} := lim_{n \to \infty} (1 + \frac{1}{n})^{n} = b_{n} := lim_{n \to \infty} \sum_{i = 0}^{n} \frac{1}{i!} \end{matrix}

右边那个单调有界都是好证的.证相等:

\begin{matrix} a_{n} = (1 + \frac{1}{n})^{n} = \sum_{i = 0}^{n} \frac{1}{n^{i}} (\binom{n}{i}) \\ = \sum_{i = 0}^{n} \frac{1}{i!} \prod_{j = 0}^{i - 1} (1 - \frac{j}{n}) < b_{n} \end{matrix}

又有

\begin{matrix} k < n \Rightarrow a_{n} = \sum_{i = 0}^{n} \frac{1}{i!} \prod_{j = 0}^{i - 1} (1 - \frac{j}{n}) > \sum_{i = 0}^{k} \frac{1}{i!} \prod_{j = 0}^{i - 1} (1 - \frac{j}{n}) \end{matrix}

于是先令 $k \to \infty$ 得到 $a_{n} > b_{k}$ ,则 $a$ 逐项大于 $b$ 的一个子列,最后得证.

重点是把一次到极限拆成两个取极限.但这个为什么对呢.

是好证的,其实就是

[think]

\begin{matrix} lim_{a \to \infty} lim_{b \to \infty} f (a, b) = X \Rightarrow lim_{n \to \infty} f (n, n) = X \end{matrix}

用定义是显然的.且如果极限都存在(要 $f (a, n)$ 和 $f (n, b)$ 极限存在)的话你还可以看出左式两取极限交换(都是右式)

\begin{matrix} 0 < e - \sum_{i = 0}^{n} \frac{1}{i!} < \frac{1}{n \cdot n!} \end{matrix}

\begin{matrix} b_{n} := \sum_{i = 0}^{n} \frac{1}{i!} \\ X = b_{n + m} - b_{n} = \sum_{i = n + 1}^{n + m} \frac{1}{i!} \\ < \frac{1}{(n + 1)!} \sum_{i = 0}^{m - 1} \frac{1}{(n + 1)^{i}} < \frac{1}{n! \cdot n} \\ \Rightarrow e - b_{n} = lim_{m \to \infty} X < \frac{1}{n! \cdot n} \end{matrix}

然后说为什么取极限小于号还是小于号呢?

\begin{matrix} X = b_{n + m} - b_{n} < b_{n + m + k} - b_{n} < \frac{1}{n! \cdot n} \\ \overset{lim_{k \to \infty}}{⟹} e - b_{n} < \frac{1}{n! \cdot n} \end{matrix}

[think] 对单调数列让一个独立变量趋近无穷说明严格不等号.

有个魔怔法,你先去下一条证明 $e$ 是无理数再回来说取不了等.

$e \notin Q$

假设 $e = \frac{p}{q}$

\begin{matrix} \frac{p}{q} - b_{n} < \frac{1}{n \cdot n!} \overset{n = q}{⟹} p (q - 1)! - b_{q} q! < \frac{1}{q} \end{matrix}

左边是不为 $0$ 的整数(为 $0$ 与 $b$ 增矛盾),右边是分数

[think] 一个无理数等价于存在一列分母到无穷的有理数始终有更高阶(相比序列中分母)的余项.

\begin{matrix} (1 + \frac{1}{n})^{n} ↑ \\ (1 + \frac{1}{n})^{n + 1} ↓ \end{matrix}

\begin{matrix} (1 + \frac{1}{n})^{n} \cdot 1 < (\frac{n \cdot (1 + \frac{1}{n} + 1)}{n + 1})^{n + 1} = (1 + \frac{1}{n + 1})^{n + 1} \end{matrix}

第二个取倒数同理.

从这个出发可以说明 $\ln$ 的切线放缩系列不等式.

有界数列必有单调子列

有界数列有收敛子列

可以考虑后缀max,取出一个单调子列.

也可以区间套,每次进有无穷多项的区间.

Class 6

Cauchy Convergance Theorem

\begin{matrix} \forall ϵ, \exists N \\ s . t . \\ \forall n, m > N, | a_{n} - a_{m} | < ϵ \Leftrightarrow lim_{n \to \infty} a_{n} exists \end{matrix}

右推左是显然的.

首先容易得到有界.于是它有收敛子列.

然后你其他的项到你的收敛子列的距离拿柯西的条件放缩一下就证完了.

或者也可以闭区间套.

确界原理

有上界的数集一定有上确界.

来闭区间套,二等分,如果上面的(包含边界)有就取上面,否则取下面,框出一个数.

然后来看,比他小的不是上界是好说的(取个区间即可).怎么说明它是上界呢?

考虑任何一个数,递归后一定有某一次它在下半区间(包含中点),那就证完了.

可能甚至不如无穷小数简洁反正本质相同.

有限覆盖定理

\begin{matrix} S = {(x, y) | x < y}, T \subset S, \cup_{I \in T} I \supset [a, b] \\ \Rightarrow \exists A \in T, \cup_{I \in A} I \supset [a, b], | A | \in N (not infinity) \end{matrix}

反证,设 $[A, B]$ 不能被有限覆盖.

闭区间套二等分则一定有一半区间也是不能被有限覆盖的.递归到不能被有限覆盖的区间,最后弄出一个数.

但包含这个数的极小区间显然可以被有限覆盖.矛盾,得证.

很棒的啊,它完全不关心你无穷覆盖的结构而是到数的结构去了.

Class 7

Funciton’s Limits

\begin{matrix} lim_{x \to x_{0}} f (x) = A \\ \Leftrightarrow \forall ϵ, \exists δ, \\ s . t . \\ \forall x, | x - x_{0} | \in (0, δ), | f (x) - A | < ϵ \end{matrix}

Heine Theorem

\begin{matrix} lim_{x \to x_{0}} f (x) = A \\ \Leftrightarrow \forall {x_{n}}, lim_{n \to \infty} x_{n} = x_{0} \\ lim_{n \to \infty} f (x_{n}) = A \end{matrix}

正向是显然的.

反向你就反证,然后极限不存在就翻译成

\exists ϵ, \forall δ, \exists x, | x - x_{0} | | f (x) - A | \geq ϵ

于是你取一个极限是 $0$ 的 $δ$ ,得到一列收敛到 $x_{0}$ 的 $x$ ,然后用这个数列就矛盾了.

然后你可以用这种方法,把函数极限的各种性质转化到数列极限,四则运算,夹逼等.

柯西收敛

\begin{matrix} lim_{x \to x_{0}} exists \Leftrightarrow \forall ϵ, \exists δ, \forall x_{1}, x_{2} \in N^{*} (x_{0}, δ), | f (x_{1}) - f (x_{2}) | < ϵ \end{matrix}

正向是显然的.

反向的话你可以用上面Heine转化成数列,则你只需要证所有这样的数列极限相等.

然后你发现你直接取任意两个数列,然后插(奇数偶数项分别放两个数列的元素)就可以直接证明这两个数列极限相等.于是结束.

\begin{matrix} \begin{array}{ll} lim_{x \to x_{0}} f (x) = A \\ lim_{t \to t_{0}} g (t) = x_{0} \\ \exists η > 0, t \in N (t_{0}, η) \Rightarrow g (t) \neq 0 \end{array}} \\ \Rightarrow lim_{t \to t_{0}} f (g (t)) = A \end{matrix}

直接翻译成 $ϵ - δ$ 是显然的.

\begin{matrix} lim_{\to_{0}} \frac{\sin (x)}{x} = 1 \end{matrix}

Class Unknown

Continuous function’s maximum/minimum

{\begin{cases} x \in [a, b], f (x) < M \\ \forall M^{'} < M, \exists x s . t . f (x) > M^{'} \\ f is continuous \end{cases} \Rightarrow \exists x, f (x) = M

考虑取任意 $M_{1} < M$ ,可以得到一个 $f (x_{1}) \in (M_{1}, M)$ ,取 $M_{n} > f (x_{n - 1})$ 可得 $x_{n} \in (M_{n}, M)$ ,显然 $x_{n}$ 有界可以取收敛子列 $y_{n}$ ,则 $lim_{n \to \infty} f (y_{n}) = f (lim_{n \to \infty} y_{n})$ ,则于是得证

Intermediate value theorem

{\begin{cases} x \in [a, b], f is continuous \\ y \in [f (a), f (b)] \end{cases} \Rightarrow \exists x_{0}, f (x_{0}) = y

不妨设 $f (a) \leq y \leq f (b)$ , $=$ 情况显然,只考虑 $f (a) < y < f (b)$ .

取 $A = {x | f (x) < y}$ 则它有上确界 $x_{1}$ .

那么一定存在一个收敛到 $x_{1}$ 的数列 ${z_{n}}$ 你就直接发现 $lim_{n \to \infty} f (z_{n}) = f (lim_{n \to \infty} z_{n})$

$Q.E.D$

About Periodicty

{\begin{cases} f (x) is a periodic function that isn’t constant \\ f (x) is continuous \end{cases} \Rightarrow f (x) has min positive period

\begin{matrix} T := {t | t > 0, f (x + t) = f (x)} \\ t_{0} := min T \end{matrix}

若 $t_{0} \neq 0$ ,取 ${t_{n}}, t_{i} \in T, lim_{n \to \infty} t_{n} = t_{0}$ ,则 $f (x) = f (x + t_{n}) = lim_{n \to \infty} f (x + t_{n}) = f (x + lim_{n \to \infty} t_{n}) = f (x + t_{0})$

若 $t_{0} = 0$ ,因为不是常函数, 则 $\exists x_{0} s . t . f (x_{0}) \neq f (0)$ ,因为连续,和极限有保号性你可以说 $\forall x \in O (x_{0}, δ), f (x) \neq f (0)$ ,但由 $t_{0} = 0$ ,你可以找到 $t < δ$ ,那就可以再找到 $N$ 使得 $N t \in (x_{0} - δ, x_{0} + δ)$ , $N t$ 一定是周期,但 $f (0 + N t) \neq f (0)$ ,矛盾.

于是得证.

Hello World

Posted on 2025-09-02 Edited on 2025-09-16

Hello World

Introduction

上大学了从CS转向Math

长期更新数学/大学课程/生活相关内容

Function Test

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